10. Convergence of Positive Series

Exercises

On these exercises, it is very important you do the Review Exercises, because they do not tell you in advance which Convergence Test to use.

a. Overview

Find the sum of the Series: \[ S=3+2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\cdots \]

All but the first term belong to a geometric series. Find the sum of the geometric series and add in the first term.

\(S=7\)

The series can be written in summation notation as: \[ S=3+\sum_{n=1}^\infty \dfrac{4}{2^n} \] The sumation is a geometric series whose sum is \(\dfrac{a}{1-r}\), where \(a=2\) and \(r=\dfrac{1}{2}\). So the sum of the series is: \[ S=3+\dfrac{2}{1-\dfrac{1}{2}}=7 \]

dd
Using the \(n^\text{th}\)-Term Divergence Test, determine if this series diverges or say the test is inconclusive: \[ S=1+\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+\cdots \]

The series can be written in summation notation as: \[ S=1+\sum_{n=1}^\infty \dfrac{n}{n+1} \]

The series diverges by the \(n^\text{th}\)-Term Divergence Test.

The series can be written in summation notation as: \[ S=1+\sum_{n=1}^\infty \dfrac{n}{n+1} \] The limit of the \(n^\text{th}\)-term is: \[ \lim_{n\to\infty}\dfrac{n}{n+1}=1 \] Since \(1\neq0\), the series diverges by the \(n^\text{th}\)-Term Divergence Test.

dd
Using the \(n^\text{th}\)-Term Divergence Test, determine if this series diverges or say the test is inconclusive: \[ S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\cdots \]

The series can be written in summation notation as: \[ S=\sum_{n=1}^\infty \dfrac{1}{n} \]

The \(n^\text{th}\)-Term Divergence Test is inconclusive.

The series can be written in summation notation as: \[ S=\sum_{n=1}^\infty \dfrac{1}{n} \] The limit of the \(n^\text{th}\)-term is: \[\lim_{n\to\infty}\dfrac{1}{n}=0\] Since the limit is \(0\), the \(n^\text{th}\)-Term Divergence Test is inconclusive.

dd

b. The Integral Test and \(p\)-Series

Use the Integral Test to determine convergence or divergence of each series.

\(\displaystyle \sum_{n=1}^\infty ne^{-n^2}\)

Write the terms as the values of a continuous function of \(x\). Then take the integral as \(x\) goes to infinity.

\(\displaystyle \sum_{n=1}^\infty ne^{-n^2}\) converges since \(\displaystyle \int_1^\infty xe^{-x^2}dx\) is finite.

The function \(f(x)=xe^{-x^2}\) is continuous, positive and decreasing and \[ \int_1^\infty xe^{-x^2}\,dx=\left.-\,\dfrac{1}{2}e^{-x^2}\right|_1^\infty =\lim_{b\to\infty}\left[-\,\dfrac{1}{2}e^{-b^2}\right]-\left[-\,\dfrac{1}{2}e^{-1}\right] =\dfrac{1}{2e} \] Since the integral is finite, \(\displaystyle \sum_{n=1}^\infty ne^{-n^2}\) is convergent.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{\sqrt{2n^2-1}}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{\sqrt{2n^2-1}}\) diverges since \(\displaystyle \int_1^\infty (2x^2-1)^{-1/2}xdx\) is infinite.

The function \(f(x)=\dfrac{x}{\sqrt{2x^2-1}}\) is continuous, positive and decreasing and \[\begin{aligned} \int_1^\infty &\dfrac{x}{\sqrt{2x^2-1}}\,dx =\left.\dfrac{\sqrt{2x^2-1}}{2}\right|_1^\infty \\ &=\lim_{b\to\infty}\left[\dfrac{\sqrt{2b^2-1}}{2}\right] -\left[\dfrac{\sqrt{3}}{2}\right] =\infty \end{aligned}\] Since the integral is infinite, \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{\sqrt{2n^2-1}}\) is divergent.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{(n+1)\ln(n+1)}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{(n+1)\ln(n+1)}\) diverges since \(\displaystyle \int_1^\infty \dfrac{1}{(x+1)\ln(x+1)}\) is infinite.

The function \(f(x)=\dfrac{1}{(x+1)\ln(x+1)}\) is continuous, positive and decreasing and \[\begin{aligned} \int_1^\infty \dfrac{1}{(x+1)\ln(x+1)}\,dx &=\left.\ln(\ln(x+1))\right|_1^\infty \\ &=\lim_{b\to\infty}\left[\ln(\ln(b+1))\right]-\left[\ln(\ln(2))\right] =\infty \end{aligned}\] Since the integral is infinite, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{(n+1)\ln(n+1)}\) is divergent.

dd
\(\displaystyle \dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{17}+\dfrac{1}{26}+...\)

To find a formula for the general term, answer the question: "What is one less than each of the demoninators?""

The series converges since \(\displaystyle \int_1^\infty \dfrac{1}{x^2+1}dx\) is finite.

The series can be written in summation notation as: \[ \sum_{n=1}^\infty \dfrac{1}{n^2+1} \] The corresponding integral is: \[\begin{aligned} \int_1^\infty &\dfrac{1}{x^2+1}dx =\lim_{n\to\infty}\int_1^n \dfrac{1}{x^2+1}dx \\ &=\lim_{n\to\infty}\arctan n-\arctan 1 =\dfrac{\pi}{2}-\dfrac{\pi}{4} =\dfrac{\pi}{4} \end{aligned}\] Since the integral is finite, the series converges

dd

For each series, determine whether the Integral Test applies. If so, use it to determine convergence or divergence of the series.

\(\displaystyle \sum_{n=1}^\infty ne^{-n}\)

Write the terms of the series as the values of a function of \(x\). Then check \(3\) conditions:
1. \(f(x)\) is positive for \(x\) greater than some number.
2. \(f(x)\) is continuous for \(x\) greater than some number.
3. \(f(x)\) is decreasing for \(x\) greater than some number. If needed, check the derivative is negative.
If the function is continous, positive, and decreasing, then the Integral Test applies and you can take the integral to determine the convergence of the series.

The Integral Test applies. \(\displaystyle \sum_{n=1}^\infty ne^{-n}\) converges since\(\displaystyle \int_1^\infty xe^{-x}\,dx\) is finite.

We test the function \(f(x)=xe^{-x}\):
1. \(f(x)\) is positive valued for \(x \gt 0\).
2. \(f(x)\) is continuous for \(x \gt 0\).
3. \(f(x)\) is decreasing for \(x \gt 1\) since \(f'(x)=e^{-x}(1-x)\) which is negative for \(x \gt 1\).
Therefore, the Integral Test applies. Integration by parts yields: \[\begin{aligned} \int_1^\infty &xe^{-x}\,dx =\left.-e^{-x}(x+1)\right|_1^\infty \\ &=\lim_{b\to\infty}\left[-e^{-b}(b+1)\right]-\left[-e^{-1}(2)\right] =\dfrac{2}{e} \end{aligned}\] Since the integral is finite, \(\displaystyle \sum_{n=1}^\infty ne^{-n}\) is convergent.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n}\)

The Integral Test applies. \(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n}\) diverges since\(\displaystyle \int_1^\infty \dfrac{\ln x}{x}\,dx\) is infinite.

Let \(f(x)=\dfrac{\ln x}{x}\).
1. \(f(x)\) is positive valued for \(x \gt 1\).
2. \(f(x)\) is continuous for \(x \gt 0\).
3. \(f(x)\) is decreasing for \(x \gt e\) since \(f'(x)=\dfrac{1-\ln x}{x^2}\) which is negative for \(x \gt e\).
Therefore, the Integral Test applies. The integral is: \[\begin{aligned} \int_1^\infty \dfrac{\ln x}{x}\,dx =\left.\dfrac{(\ln x)^2}{2}\right|_1^\infty=\infty \end{aligned}\] Since the integral is infinite, \(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n}\) is divergent.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{1+n^2}\)

The Integral Test applies. \(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{1+n^2}\) converges since\(\displaystyle \int_1^\infty \dfrac{\arctan x}{1+x^2}\,dx\) is finite.

Let \(f(x)=\dfrac{\arctan x}{1+x^2}\).
1. \(f(x)\) is positive valued for \(x \gt 0\).
2. \(f(x)\) is continuous for \(x \gt 0\).
3. \(f(x)\) is decreasing for \(x \gt 1\) since \(f'(x)=\dfrac{1-2x\arctan x}{(1+x^2)^2}\) which is negative for \(x \gt 1\).
Therefore, the Integral Test applies. The integral is: \[\begin{aligned} \int_1^\infty &\dfrac{\arctan x}{1+x^2}\,dx =\left. \dfrac{(\arctan x)^2}{2}\right|_1^\infty \\ &=\lim_{b\to\infty}\left[\dfrac{(\arctan b)^2}{2}\right] -\left[\dfrac{(\arctan 1)^2}{2}\right] \\ &=\dfrac{1}{2}\left[\left(\dfrac{\pi}{2}\right)^2-\left(\dfrac{\pi}{4}\right)^2\right] =\dfrac{3\pi^2}{32} \end{aligned}\] Since the integral is finite, \(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{1+n^2}\) is convergent.

dd
\(\displaystyle \sum_{n=1}^\infty (\sin n)e^{-n}\)

The Integral Test does not apply. A different test will be necessary to determine if the series converges.

We test the function \(f(x)=(\sin x)e^{-x}\):
1. \(f(x)\) is not positive valued for \(x \gt 0\) because the value of \(\sin x\) will oscilliate between-1 and 1.
2. \(f(x\) is continuous for \(x \gt 0\).
3. \(f(x)\) is not always decreasing since \(f'(x)=(\cos x)e^{-x}-(\sin x)e^{-x}\) will also oscilliate between positive and negative values.
Therefore, the Integral Test does not apply. (Condition 1 or 3 is sufficient.) A different test will be necessary to determine if the series converges.

dd
\(\displaystyle \sum_{n=1}^\infty (2+\sin n)e^{-n}\)

The Integral Test applies, \(\displaystyle \sum_{n=1}^\infty (2+\sin n)e^{-n}\) converges since \(\displaystyle\int_1^\infty (2+\sin x)e^{-x}\,dx\) is finite.

We test the function \(f(x)=(2+\sin x)e^{-x}\):
1. \(f(x)\) is positive valued for all \(x\).
2. \(f(x)\) is continuous for all \(x\).
3. \(f(x)\) is decreasing for all \(x\) since \(f'(x)=e^{-x}(-2-\sin(x)+\cos(x))\) which is negative for all \(x\).
Therefore, the Integral Test applies. We split the integral: Using integration by parts yields: \[ \int_1^\infty (2+\sin x)e^{-x}\,dx =\int_1^\infty 2e^{-x}\,dx+\int_1^\infty (\sin x)e^{-x}\,dx \] The first integral is: \[ \int_1^\infty 2e^{-x}\,dx =\left[-2e^{-x}\right]_1^\infty =0--\dfrac{2}{e}=\dfrac{2}{e} \] For the second integral, we integrate by parts twice and solve for the recurring integral: \[\begin{aligned} \int_1^\infty (\sin x)e^{-x}\,dx &=\left[-\dfrac{1}{2}e^{-x}\cos x-\dfrac{1}{2}e^{-x}\sin x\right]_1^\infty \\ &=0+\dfrac{1}{2e}(\cos 1+\sin 1) \end{aligned}\] Since the total integral is finite, \(\displaystyle \sum_{n=1}^\infty (2+\sin n)e^{-n}\) is convergent.

dd

Determine whether each \(p\)-Series converges or diverges.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 4]{n^5}}\)

Convergent p-series: \(p=5/4 \gt 1\).

dd

Since \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 4]{n^5}} =\sum_{n=1}^\infty \dfrac{1}{n^{5/4}}\), this is a \(p\)-series with \(p=5/4\). Since \(p=5/4 \gt 1\), it is convergent.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{\sqrt[\scriptstyle 3]{n^4}}\)

Divergent \(p\)-series: \(p=1/3 \lt 1\).

Since \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{\sqrt[\scriptstyle 3]{n^4}} =\sum_{n=1}^\infty \dfrac{n}{n^{4/3}} =\sum_{n=1}^\infty \dfrac{1}{n^{1/3}} \), this is a \(p\)-series with \(p=1/3\). Since \(p=1/3 \lt 1\), it is divergent.

dd
\(1+\dfrac{1}{\sqrt[\scriptstyle 3]{2}} +\dfrac{1}{\sqrt[\scriptstyle 3]{3}} +\dfrac{1}{\sqrt[\scriptstyle 3]{4}}+...\)

Find a formula for the general term of the series.

Divergent \(p\)-series: \(p=1/3 \lt 1\).

The series can be written in summation notation as: \[ \sum_{n=1}^\infty \dfrac{1}{n^{1/3}} \] This is a \(p\)-series with \(p=1/3 \lt 1\). Therefore the series is divergent.

dd
\(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\dfrac{1}{4\sqrt{4}}+...\)

Convergent \(p\)-series: \(p=3/2 \gt 1\)

The series can be written in summation notation as: \[ \sum_{n=1}^\infty \dfrac{1}{n\sqrt{n}}= \sum_{n=1}^\infty \dfrac{1}{n^{3/2}} \] This is a \(p\)-series with \(p=3/2 \gt 1\). Therefore the series is convergent.

dd
\(3+\dfrac{3}{\sqrt[\scriptstyle 3]{4}}+\dfrac{3}{\sqrt[\scriptstyle 3]{9}} +\dfrac{3}{\sqrt[\scriptstyle 3]{16}}+\dfrac{3}{\sqrt[\scriptstyle 3]{25}}+...\)

Divergent \(p\)-series: \(p=2/3 \lt 1\)

The series can be written in summation notation as: \[ \sum_{n=1}^\infty \dfrac{3}{n^{2/3}} \] This is a \(p\)-series with \(p=2/3 \lt 1\). Therefore the series is divergent.

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c. The Simple Comparison Test

Use the Simple Comparison Test with a known series to determine the convergence or divergence of each series.

Be careful you don't confuse a \(p\)-series, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p}\), with a geometric series, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{p^n}\). In each case, \(p\) is the constant.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n^3+1}}\)

Remember, the larger the demoninator, the smaller the fraction.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n^3+1}}\) converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n^3}}\).

Since \(\displaystyle \dfrac{1}{\sqrt{n^3+1}}<\dfrac{1}{\sqrt{n^3}}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n^3}}\) is a convergent \(p\)-series (\(p=\dfrac{3}{2}>1\)), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n^3+1}}\) also converges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{2^n-1}\)

Note: \(3\) and \(2\) are raised to the same power. To find a comparison series, drop the \(-1\) and rewrite the result as a geometric series.
Remember, the smaller the demoninator, the larger the fraction.

\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{2^n-1}\) diverges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{3}{2}\right)^n\).

Since \(\displaystyle \dfrac{3^n}{2^n-1} \gt \dfrac{3^n}{2^n} =\left(\dfrac{3}{2}\right)^n\) and \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{3}{2}\right)^n\) is a divergent geometric series \((r=\dfrac{3}{2} \gt 1)\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{2^n-1}\) also diverges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^3}\)

Note: \(\ln n \lt n\).

\(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^3}\) converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\).

Since \(\ln n< n\), we know \(\dfrac{\ln n}{n^3} \lt \dfrac{n}{n^3}=\dfrac{1}{n^2}\). Since the series \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\) is a convergent \(p\)-series (\(p=2 \gt 1\)), we conclude the series \(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^3}\) also converges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{n^2}\)

Note: \(|\cos n| \le 1\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{n^2}\) converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\).

Since \(-1 \le \cos n \le 1\), we have \(\dfrac{1}{n^2} \le \dfrac{2+\cos n}{n^2} \le \dfrac{3}{n^2}\).
Since both \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\) are convergent \(p\)-series (\(p=2 \gt 1\)), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{n^2}\) also converges.

Notice that we found both the upper bound and lower bound of the series. However, to satisfy the Simple Comparison Test for convergence, we only have to compare to the upper bound, \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\).

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{\sqrt{n}}\)

Note: \(|\cos n| \le 1\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{\sqrt{n}}\) diverges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}\).

Since \(-1 \le \cos n \le 1\), we have \(\dfrac{1}{n^2} \le \dfrac{2+\cos n}{n^2} \le \dfrac{3}{n^2}\).
Since both \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{\sqrt{n}}\) are divergent \(p\)-series (\(p=\dfrac{1}{2} \lt 1\)), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\cos n}{n^2}\) also diverges.

Notice that we found both the upper bound and lower bound of the series. However, to satisfy the Simple Comparison Test for divergence, we only have to compare to the lower bound, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}\).

dd,py
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^n}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^n}\) converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\).

Since \(\displaystyle \dfrac{1}{n^n} \le \dfrac{1}{n^2}\) for \(n \ge 1\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\) is a convergent \(p\)-series with \(p=2 \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^n}\) also converges.

dd
\(\displaystyle \sum_{n=2}^\infty \dfrac{\sqrt{n}}{n^{3/2}-1}\)

\(\displaystyle \sum_{n=2}^\infty \dfrac{\sqrt{n}}{n^{3/2}-1}\) diverges by simple comparison with the harmonic series \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\).

Since \(\displaystyle \dfrac{\sqrt{n}}{n^{3/2}-1} \gt \dfrac{\sqrt{n}}{n^{3/2}} =\dfrac{1}{n}\) and \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\) is the divergent harmonic series, the series \(\displaystyle \sum_{n=2}^\infty \dfrac{\sqrt{n}}{n^{3/2}-1}\) also diverges.

dd,py
\(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n^2+1}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n^2+1}\) converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}\).

Since \(\displaystyle \dfrac{\sqrt{n}}{n^2+1} \lt \dfrac{\sqrt{n}}{n^2} =\dfrac{1}{n^{3/2}}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}\) is a convergent \(p\)-series with \(p=\dfrac{3}{2} \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n^2+1}\) also converges.

dd
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20} +\cdots+\dfrac{1}{n(n+1)}+\cdots\)

Series converges by simple comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\).

Since \(\displaystyle \dfrac{1}{n(n+1)}<\dfrac{1}{n^2}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\) is a convergent \(p\)-series (\(p=2 \gt 1\)), this series also converges.

dd

\(\dfrac{1}{n(n+1)}\) can also be decomposed into partial fractions, giving \(\dfrac{1}{n}-\dfrac{1}{n+1}\).
Therefore, the series \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(n+1)}\) can also be written as the telescoping series: \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\). The partial sum is \[ S_k=\sum_{n=1}^k \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{k+1} \] So the sum is \(\displaystyle S=\lim_{n\to\infty}S_k=1\). Consequently, not only can we show the series is convergent, we can actually find its sum is \(1\).

d. The Limit Comparison Test

Use the Limit Comparison Test with a known series to determine convergence or divergence of each series.

\(\displaystyle \sum_{n=1}^\infty \dfrac{3n-5}{n^2-n+3}\)

To pick a series for \(b_n\) keep the largest term in the numerator and the largest term the denominator of \(a_n\).
This will ensure the limit converges.

\(\displaystyle \sum_{n=1}^\infty \dfrac{3n-5}{n^2-n+3}\) diverges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n}\).

Let \(b_n\) equal the ratio of the highest ordered terms in the numerator and denominator of \(a_n=\dfrac{3n-5}{n^2-n+3}\). Thus: \[ b_n=\dfrac{3n}{n^2}=\dfrac{3}{n} \] Then we compute: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\;\dfrac{3n-5}{n^2-n+3}\;}{\dfrac{3}{n}} =\lim_{n\to\infty}\dfrac{3n-5}{n^2-n+3}\cdot\dfrac{n}{3} \\ &=\lim_{n\to\infty}\dfrac{3n^2-5n}{3(n^2-n+3)}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n}\) is a divergent harmonic series, \(\displaystyle \sum_{n=1}^\infty \dfrac{3n-5}{n^2-n+3}\) diverges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{n^3+2n}{2^n(n^3+1)}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{n^3+2n}{2^n(n^3+1)}\) converges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n}\).

Let \(b_n=\dfrac{1}{2^n}\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\;\dfrac{n^3+2n}{2^n(n^3+1)}\;}{\dfrac{1}{2^n}} =\lim_{n\to\infty}\dfrac{n^3+2n}{2^n(n^3+1)}\cdot\dfrac{2^n}{1} \\ &=\lim_{n\to\infty}\dfrac{n^3+2n}{n^3+1}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n}\) is a convergent geometric series \((|r|=\dfrac{1}{2} \lt 1)\), \(\displaystyle \sum_{n=1}^\infty \dfrac{n^3+2n}{2^n(n^3+1)}\) converges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+2}}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+2}}\) diverges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\).

To find a comparison series, we drop the \(+2\). Thus \(b_n=\dfrac{1}{n}\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty} \dfrac{\;\dfrac{1}{\sqrt[\scriptstyle 3]{n^3+2}}\;}{\dfrac{1}{n}} =\lim_{n\to\infty} \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+2}}\cdot\dfrac{n}{1}\\[5pt] &=\lim_{n\to\infty} \dfrac{n}{\sqrt[\scriptstyle 3]{n^3+2}}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\) is the divergent harmonic series, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+2}}\) diverges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^4-n}}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^4-n}}\) converges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{4/3}}\).

To find a comparison series, we drop the \(-n\). Thus \(b_n=\dfrac{1}{n^{4/3}}\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty} \dfrac{\;\dfrac{1}{\sqrt[\scriptstyle 3]{n^4-n}}\;}{\dfrac{1}{n^{4/3}}} =\lim_{n\to\infty}\dfrac{1}{\sqrt[\scriptstyle 3]{n^4-n}} \cdot\dfrac{n^{4/3}}{1}\\ &=\lim_{n\to\infty}\left(\dfrac{n^4}{n^4-n}\right)^{1/3}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{4/3}}\) is a convergent \(p\)-series (\(p=\dfrac{4}{3} \gt 1\)), \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^4-n}}\) converges.

dd
\(\displaystyle \sum_{n=1}^\infty \dfrac{2+2^n}{3+3^n}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{2+2^n}{3+3^n}\) converges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{2}{3}\right)^n\).

To find a comparison series, we drop the \(2\) in the numerator and the \(3\) in the denominator. Thus \(b_n=\dfrac{2^n}{3^n}\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\;\dfrac{2+2^n}{3+3^n}\;}{\dfrac{2^n}{3^n}} =\lim_{n\to\infty}\dfrac{2+2^n}{3+3^n}\cdot\dfrac{2^{-n}}{3^{-n}}\\ &=\lim_{n\to\infty} \dfrac{2^{1-n}+1}{3^{1-n}+1}=1 \end{aligned}\] Notice that the \(2^{1-n}\) and \(3^{1-n}\) terms go to \(0\) as \(n\to\infty\). Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{2}{3}\right)^n\) is a convergent geometric series (\(|r|=\dfrac{2}{3} \lt 1\)), \(\displaystyle \sum_{n=1}^\infty \dfrac{2+2^n}{3+3^n}\) converges.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n+4}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n+4}\) diverges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}\).

To find a comparison series, we drop the \(+4\). Thus \(b_n=\dfrac{\sqrt{n}}{n}=\dfrac{1}{\sqrt{n}}\). We compute: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\;\dfrac{\sqrt{n}}{n+4}\;}{\dfrac{1}{\sqrt{n}}} =\lim_{n\to\infty}\dfrac{\sqrt{n}}{n+4}\cdot\sqrt{n}\\ &=\lim_{n\to\infty}\dfrac{n}{n+4}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{1/2}}\) is a divergent \(p\)-series (\(p=\dfrac{1}{2} \lt 1\)), \(\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{n}}{n+4}\) diverges.

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\(\displaystyle \sum_{n=1}^\infty \sin\left(\dfrac{1}{n^2}\right)\)

\(\displaystyle \sum_{n=1}^\infty \sin\left(\dfrac{1}{n^2}\right)\) converges by a limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\).

To find a comparison series, we recall that the linear approximation to \(\sin x\) is \(x\). Thus, \(\sin\left(\dfrac{1}{n^2}\right)\approx\dfrac{1}{n^2}\). So we take \(b_n=\dfrac{1}{n^2}\). Then: \[ \lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{\;\sin\dfrac{1}{n^2}\;}{\dfrac{1}{n^2}} \] Let \(t=\dfrac{1}{n^2}\). Then \(n\to\infty\) means \(t\to0^+\) and so: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{t\to0^+}\dfrac{\sin t}{t}=1 \end{aligned}\] which is a standard limit from single variable calculus. Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}\) is a convergent \(p\)-series (\(p=2 \gt 1\)), \(\displaystyle \sum_{n=1}^\infty \sin\left(\dfrac{1}{n^2}\right)\) also converges.

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\(\displaystyle \dfrac{1}{2}+\dfrac{2}{5}+\dfrac{3}{10}+\dfrac{4}{17}+\dfrac{5}{26}+...\)

Find the general term. (What is one less than each of the denominators?)

The series diverges by limit comparison with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\).

The general term is \(\dfrac{n}{n^2+1}\). Therefore, let \(b_n=\dfrac{1}{n}\). So: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\;\dfrac{n}{n^2+1}\;}{\dfrac{1}{n}} =\lim_{n\to\infty}\dfrac{n}{n^2+1}\cdot\dfrac{n}{1}\\ &=\lim_{n\to\infty}\dfrac{n^2}{n^2+1}=1 \end{aligned}\] Therefore, since \(0 < 1 < \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\) is the divergent harmonic series, \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{n^2+1}\) diverges.

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\(\displaystyle \sum_{n=2}^\infty \dfrac{\ln n}{n^{3/2}}\)

\(\displaystyle \sum_{n=2}^\infty \dfrac{\ln n}{n^{3/2}}\) converges by a limit comparison with \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{5/4}}\).

It appears we should compare to a \(p\)-series, but for what \(p\)? Since there is an \(n^{3/2}\) in the denominator, we first try a comparison to the \(p\)-series \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{3/2}}\) which is convergent since \(p=\dfrac{3}{2} \gt 1\). Then: \[ L=\lim_{n\to\infty} \dfrac{a_n}{b_n} =\lim_{n\to\infty} \dfrac{\;\dfrac{\ln n}{n^{3/2}}\;}{\dfrac{1}{n^{3/2}}} =\lim_{n\to\infty} \ln n=\infty \] This is one of the extreme cases. Since \(L=\infty\) but \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{3/2}}\) is convergent, the Limit Comparison Test FAILS. So maybe it diverges.

So we next try a comparison to the divergent harmonic series \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\). Then: \[\begin{aligned} L&=\lim_{n\to\infty} \dfrac{\;\dfrac{\ln n}{n^{3/2}}\;}{\dfrac{1}{n}} =\lim_{n\to\infty} \dfrac{\ln n}{n^{1/2}} \\ &\,\overset{\text{l'H}}{=}\,\lim_{n\to\infty} \dfrac{\;\dfrac{1}{n}\;}{\dfrac{1}{2n^{1/2}}} =\dfrac{2}{n^{1/2}}=0 \end{aligned}\] This is again an extreme case. Since \(L=0\) but \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\) is divergent, the Limit Comparison Test again FAILS. So maybe we need something between \(\dfrac{1}{n^{3/2}}\) and \(\dfrac{1}{n}\).

So we now try a comparison to \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{5/4}}\) which is convergent since \(p=\dfrac{5}{4} \gt 1\). Then: \[\begin{aligned} L&=\lim_{n\to\infty} \dfrac{\;\dfrac{\ln n}{n^{3/2}}\;}{\dfrac{1}{n^{5/4}}} =\lim_{n\to\infty} \dfrac{\ln n}{n^{1/4}} \\ &\,\overset{\text{l'H}}{=}\,\lim_{n\to\infty} \dfrac{\;\dfrac{1}{n}\;}{\dfrac{1}{4n^{3/4}}} =\dfrac{4}{n^{1/4}}=0 \end{aligned}\] Again an extreme case. However, since \(L=0\) and \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{5/4}}\) is convergent, the Limit Comparison Test says \(\displaystyle \sum_{n=2}^\infty \dfrac{\ln n}{n^{3/2}}\) also converges.

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In a Math paper, the mathematician would only give the last \(\dfrac{1}{3}\) of this solution and the reader is left wondering how they came up with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{5/4}}\). The answer is, the mathematician went through the first \(\dfrac{2}{3}\) of this solution but did not write it in the paper.
\(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n\ln n}\)

\(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n\ln n}\) diverges by the Integral Test.

It appears we should compare to a \(p\)-series, but for what \(p\)? Since there is an \(n\) in the denominator, we first try a comparison to the divergent harmonic series \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\). Then: \[ L=\lim_{n\to\infty} \dfrac{\;\dfrac{1}{n\ln n}\;}{\dfrac{1}{n}} =\lim_{n\to\infty} \dfrac{1}{\ln n}=0 \] This is one of the extreme cases. Since \(L=0\) but \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\) is divergent, the Limit Comparison Test again FAILS. So maybe it is convergent.

So we now try a comparison to \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^p}\) with \(p \gt 1\) so it will be convergent. Then: \[\begin{aligned} L&=\lim_{n\to\infty} \dfrac{\;\dfrac{1}{n\ln n}\;}{\dfrac{1}{n^p}} =\lim_{n\to\infty} \dfrac{n^{p-1}}{\ln n} \\ &\,\overset{\text{l'H}}{=}\,\lim_{n\to\infty} \dfrac{\;(p-1)n^{p-2}\;}{\dfrac{1}{n}} =\lim_{n\to\infty} (p-1)n^{p-1}=\infty \end{aligned}\] This is again an extreme case. Since \(L=\infty\) but \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p}\) is convergent, the Limit Comparison Test again FAILS.

So what do we do? We tried \(p=1\) and \(p \gt 1\). If you try \(p \lt 1\), it will also FAIL. Try it! What do we do? We go back to first principles. Notice that \(f(x)=\dfrac{1}{x\ln x}\) is a contuous, positive, decreasing function. So we can apply the Integral Test. We will need the substitution \(u=\ln x\) with \(du=\dfrac{1}{x}\,dx\): \[\begin{aligned} \int_1^\infty &\dfrac{1}{x\ln x}\,dx =\int_{x=1}^\infty \dfrac{1}{u}\,du =\left[\ln u\rule{0pt}{10pt}\right]_{x=1}^\infty \\ &=\left[\ln\ln x\rule{0pt}{10pt}\right]_{x=1}^\infty =\infty--\infty=\infty \end{aligned}\] Since the integral is divergent, so is the series \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n\ln n}\).

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e. The Ratio Test

Use the Ratio Test to test each series for convergence. (You will get more practice with the Ratio Tests in the next chapter).

\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n+2}{4^n}\)

For the Ratio Test, compute the limit of the ratio of seccessive terms: \[ \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} \]

\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n+2}{4^n}\) converges by the Ratio Test.

We have \(a_n=\dfrac{3^n+2}{4^n}\) and \(a_{n+1}=\dfrac{3^{n+1}+2}{4^{n+1}}\). We compute the limit of the ratio of seccessive terms: \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}= \lim_{n\to\infty}\dfrac{3^{n+1}+2}{4^{n+1}}\cdot\dfrac{4^n}{3^n+2}\\ &=\lim_{n\to\infty}\dfrac{3^{n+1}+2}{4(3^n+2)} =\dfrac{3}{4} \end{aligned}\] (Notice how we divide by \(a_n\) by multiplying by its reciprocal.) Since \(\rho=\dfrac{3}{4}< 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{3^n+2}{4^n}\) converges by the Ratio Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{n(n+5)}{n!}\)

Remember: \[ \dfrac{n!}{(n+1)!} =\dfrac{n(n-1)\cdots1}{(n+1)n(n-1)\cdots1} =\dfrac{1}{n+1} \]

\(\displaystyle \sum_{n=1}^\infty \dfrac{n(n+5)}{n!}\) converges by the Ratio Test.

We have \(a_n=\dfrac{n(n+5)}{n!}\) and \(a_{n+1}=\dfrac{(n+1)(n+6)}{(n+1)!}\). The limit of the ratio of succesive terms is: \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}= \lim_{n\to\infty}\dfrac{(n+1)(n+6)}{(n+1)!}\cdot\dfrac{n!}{n(n+5)}\\ &=\lim_{n\to\infty} \dfrac{(n+1)(n+6)}{(n+1)n(n+5)} =\lim_{n\to\infty} \dfrac{(n+6)}{(n^2+5n)}=0 \end{aligned}\] Since \(\rho=0< 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{n(n+5)}{n!}\) converges by the Ratio Test.

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\(\displaystyle \sum_{n=1}^\infty n!e^{-n}\)

\(\displaystyle \sum_{n=1}^\infty n!e^{-n}\) diverges by the Ratio Test.

We have \(a_n=n!e^{-n}\) and \(a_{n+1}=(n+1)!e^{-n-1}\). The limit of the ratio of succesive terms is: \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{(n+1)!e^{-n-1}}{n!e^{-n}}\\ &=\lim_{n\to\infty}\dfrac{n+1}{e} =\infty \end{aligned}\] Since \(\rho=\infty \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty n!e^{-n}\) diverges by the Ratio Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{n!}{10^{2n}}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{n!}{10^{2n}}\) diverges by the Ratio Test.

We have \(a_n=\dfrac{n!}{10^{2n}}\) and \(a_{n+1}=\dfrac{(n+1)!}{10^{2(n+1)}}\). The limit of the ratio of succesive terms is: \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}= \lim_{n\to\infty}\dfrac{(n+1)!}{10^{2n+2}}\cdot\dfrac{10^{2n}}{n!}\\ &=\lim_{n\to\infty}\dfrac{(n+1)}{10^2} =\infty \end{aligned}\] Since \(\rho=\infty \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{n!}{10^{2n}}\) diverges diverges by the Ratio Test.

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Display enough terms of the following recursively defined series to be able to generalize to an expression for \(a_n\). Then use the Ratio Test to determine convergence or divergence. Finally, explain why you did not really need to find a general expression for \(a_n\).

\(\displaystyle \sum_{n=1}^\infty a_n\) where \(a_1=1\) and \(a_{n+1}=\dfrac{2}{n}a_n\)

Write out the first 6 terms of the sequence without mutlplying out the results (leave the terms unsimpflied). This will help you notice the pattern in the sequence of terms so you can generalize it.

The series converges by the RatioTest.

\[\begin{aligned} a_1&=1 \qquad \text{and} \qquad a_{n+1}=\dfrac{2}{n}a_n \\ a_2&=\dfrac{2}{1}\cdot1 \\ a_3&=\dfrac{2}{2}\dfrac{2}{1}\cdot1 \\ a_4&=\dfrac{2}{3}\dfrac{2}{2}\dfrac{2}{1}\cdot1 \\ a_5&=\dfrac{2}{4}\dfrac{2}{3}\dfrac{2}{2}\dfrac{2}{1}\cdot1 \\ a_6&=\dfrac{2}{5}\dfrac{2}{4}\dfrac{2}{3}\dfrac{2}{2}\dfrac{2}{1}\cdot1 \\ &\vdots \\ a_n&=\dfrac{2^{n-1}}{(n-1)!} \end{aligned}\] Then, using the Ratio Test: \[ \lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty} \dfrac{2^n}{n!}\cdot\dfrac{(n-1)!}{2^{n-1}} =\lim_{n\to\infty} \dfrac{2}{n} =0 \lt 1 \] So the series converges by the RatioTest.

Finally, notice that we did not really need to find a general expression for \(a_n\). The recursion formula, \(a_{n+1}=\dfrac{2}{n}a_n\), gives directly: \[ \lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty} \dfrac{2}{n}=0 \]

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f. The Root Test

Use the Root Test to test each series for convergence. (You will get more practice with the Root Tests in the next chapter).

\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{2^n}\)

For the Root Test, compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[ \rho=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} \]

To compute the limit, recall \(\displaystyle \lim_{n\to\infty}\sqrt[\scriptstyle n]{n}=1\) because: \[\begin{aligned} \lim_{n\to\infty} &\sqrt[\scriptstyle n]{n} ={\large e}^{\displaystyle \,\lim_{n\to\infty}\ln(n^{1/n})} \\ &={\large e}^{\displaystyle \,\lim_{n\to\infty} \scriptstyle\frac{\scriptstyle\ln(n)}{\scriptstyle n}} ={\large e}^0 =1 \end{aligned}\] Notice this is a special case of the Limit of the \(n^\text{th}\) Root of a Polynomial Proposition

\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{2^n}\) converges by the Root Test.

We compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} =\lim_{n\to\infty}\left(\dfrac{n}{2^n}\right)^{1/n} \\ &=\lim_{n\to\infty}\dfrac{\sqrt[\scriptstyle n]{n}}{2} =\dfrac{1}{2} \end{aligned}\] (See the hint for why \(\displaystyle \lim_{n\to\infty}\sqrt[\scriptstyle n]{n}=1\).)
Since \(\rho=\dfrac{1}{2}<1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{2^n}\) converges by the Root Test.

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\(\displaystyle \sum_{n=1}^\infty n^2e^{-n}\)

\(\displaystyle \sum_{n=1}^\infty n^2e^{-n}\) converges by the Root Test.

We compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} =\lim_{n\to\infty}\left(\dfrac{n^2}{e^n}\right)^{1/n} \\ &=\lim_{n\to\infty}\dfrac{(n^{1/n})^2}{e}=\dfrac{1}{e} \end{aligned}\] Since \(\rho=\dfrac{1}{e}<1\), the series \(\displaystyle \sum_{n=1}^\infty n^2e^{-n}\) converges by the Root Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{n2^n}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{n2^n}\) diverges by the Root Test.

We compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} =\lim_{n\to\infty}\left(\dfrac{3^n}{n2^n}\right)^{1/n} \\ &=\lim_{n\to\infty}\dfrac{3}{\sqrt[\scriptstyle n]{n}2}=\dfrac{3}{2} \end{aligned}\] Since \(\rho=\dfrac{3}{2}>1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{3^n}{n2^n}\) diverges by the Root Test.

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\(\displaystyle \sum_{n=2}^\infty \dfrac{n}{(\ln n)^n}\)

\(\displaystyle \sum_{n=2}^\infty \dfrac{n}{(\ln n)^n}\) converges by the Root Test.

We compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} =\lim_{n\to\infty}\left(\dfrac{n}{(\ln n)^n}\right)^{1/n} \\ &=\lim_{n\to\infty}\dfrac{\sqrt[\scriptstyle n]{n}}{\ln n} =``\dfrac{1}{\infty}" =0 \end{aligned}\] Since \(\rho=0 \lt 1\), the series \(\displaystyle \sum_{n=2}^\infty \dfrac{n}{(\ln n)^n}\) converges by the Root Test.

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\(\displaystyle \sum_{n=2}^\infty \dfrac{n^4+n^3+3n+3}{n^{n+1}+n^n}\)

Remember the Limit of the \(n^\text{th}\) Root of a Polynomial Proposition: \[ \lim_{n\to\infty}\sqrt[\scriptstyle n]{p(n)}=1 \]

\(\displaystyle \sum_{n=2}^\infty \dfrac{n^4+n^3+3n+3}{n^{n+1}+n^n}\) converges by the Root Test.

We compute the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term: \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\scriptstyle n]{a_n} =\lim_{n\to\infty}\left(\dfrac{n^4+n^3+3n+3}{n^{n+1}+n^n}\right)^{1/n} \\ &=\lim_{n\to\infty}\left(\dfrac{n^4+n^3+3n+3}{n^n(n+1)}\right)^{1/n} \\ &=\lim_{n\to\infty}\dfrac{1}{n}\dfrac{\left(n^4+n^3+3n+3\right)^{1/n}}{\left(n+1\right)^{1/n}} \\ &=0\cdot\dfrac{1}{1}=0 \end{aligned}\] Since \(\rho=0 < 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{n^4+n^3+3n+3}{n^{n+1}+n^n}\) converges by the Root Test.

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>

g. Approximating the Series

If each of the follow series is approximated by its \(50^\text{th}\)-partial sum, use an integral to find an upper bound on the error in the approximation.

\(\displaystyle \sum_{n=1}^\infty e^{-2n}\)

\(E_{50}<\dfrac{1}{2}e^{-100}\approx 1.86\times10^{-44}\)

The error \(E_{50}\) is bounded by \[ \int_{50}^\infty e^{-2n}\,dn =\left[\dfrac{e^{-2n}}{-2}\right]_{50}^\infty =\dfrac{1}{2}e^{-100} \approx 1.86\times10^{-44} \]
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n}\)

Recall: \(\displaystyle \int b^x \,dx=\dfrac{b^x}{\ln b}+C\)

\(E_{50}<\dfrac{2^{-50}}{\ln2}\approx 1.3\times10^{-15}\)

The error \(E_{50}\) is bounded by \[ \int_{50}^\infty 2^{-n}\,dn =\left[\dfrac{-2^{-n}}{\ln(2)}\right]_{50}^\infty =\dfrac{2^{-50}}{\ln(2)} \approx 1.3\times10^{-15} \]
\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^3}\)

\(E_{50}<\dfrac{1}{5000}\approx 2\times10^{-4}\)

The error \(E_{50}\) is bounded by \[ \int_{50}^\infty \dfrac{1}{n^3}\,dn =\left[\dfrac{-1}{2n^2}\right]_{50}^\infty =\dfrac{1}{5000}=2\times10^{-4} \]

For each of the following series, use a Simple Comparison Test to show convergence. Then if the series is approximated by its \(5^\text{th}\)-partial sum, find an upper bound on the error in the approximation.

\(\displaystyle \sum_{n=1}^\infty \dfrac{5+\dfrac{1}{n}}{3^n}\)

\(E_{5} \lt \dfrac{1}{81}\)

Since \(5+\dfrac{1}{n} \le 6\), we compare to: \(\displaystyle \sum_{n=1}^\infty \dfrac{6}{3^n}\), which converges because it is a geometric series with ratio \(r=\dfrac{1}{3} \lt 1\). If the series is approximated by its \(5^\text{th}\)-partial sum, then the error satisfies \[ E_{5}=\sum_{n=6}^\infty \dfrac{5+\dfrac{1}{n}}{3^n} \lt \sum_{n=6}^\infty \dfrac{6}{3^n}=\dfrac{6/3^6}{1-1/3}=\dfrac{1}{81} \]

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\(\displaystyle \sum_{n=2}^\infty \dfrac{n^2-1}{n^4}\)
See the hint.

The \(5^\text{th}\)-partial sum is: \[ S_5=\sum_{n=2}^5 \dfrac{n^2-1}{n^4} \] Notice \(n^2-1 \lt n^2\).
Also, two known series are \[ \sum_{n=1}^\infty \dfrac{1}{n^2}=\dfrac{\pi^2}{6} \quad \text{and} \quad \sum_{n=1}^\infty \dfrac{1}{n^4}=\dfrac{\pi^6}{945} \]

\(E_{5} \lt 0.1813\)

Since \(n^2-1 \lt n^2\), we compare to: \(\displaystyle \sum_{n=2}^\infty \dfrac{n^2}{n^4} =\sum_{n=2}^\infty \dfrac{1}{n^2}\) which is a convergent p-series with \(p=2\). So the original series converges by the Simple Comparison Test. If the series is approximated by its \(5^\text{th}\)-partial sum, then the error satisfies: \[\begin{aligned} E_{5}&=\sum_{n=6}^\infty \dfrac{n^2-1}{n^4} \lt \sum_{n=6}^\infty \dfrac{1}{n^2} =\sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^5 \dfrac{1}{n^2} \\ &=\dfrac{\pi^2}{6}-\left(1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}\right) =0.1813 \end{aligned}\]

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\(\displaystyle \sum_{n=1}^\infty \left(\dfrac{\sin^2n}{n}-\dfrac{1}{n+1}\right)\)

Remember how to sum a telescoping series.

\(E_{5} \lt \dfrac{1}{6}\)

Since \(\sin^2 n \le 1\), we can compare to \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\). This is a telescoping series. It's \(k^\text{th}\) partial sum is \[ S_k=\left(\dfrac{1}{1}-\dfrac{1}{2}\right) +\left(\dfrac{1}{2}-\dfrac{1}{3}\right) +\cdots+\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right) \] So the series converges to \[ \sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) =\lim_{k\to\infty} \left(1-\dfrac{1}{k}\right)=1 \] So the original series converges by the Simple Comparison Test. If we approximate the series by its \(5^\text{th}\)-partial sum, then the error satisfies \[ E_{5}=\sum_{n=6}^\infty \left(\dfrac{\sin^2n}{n}-\dfrac{1}{n+1}\right) \lt \sum_{n=6}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) =\lim_{k\to\infty} \left(\dfrac{1}{6}-\dfrac{1}{k}\right)=\dfrac{1}{6} \]

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Review Exercises

Use any appropriate test to determine whether each of the following converges or diverges. The particular test cited in the answer may not be the only appropriate test for convergence! Look back at the Overview for a summary of all the tests. It is very important you do these Review Exercises, because they do not tell you in advance which Convergence Test to use.

\(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n(\ln n)^3}\)

For what \(x\)'s is the function \(f(x)=\dfrac{1}{x(\ln x)^3}\) continuous, positive and decreasing? Find the derivative to see it is decreasing.

\(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n(\ln n)^3}\) converges by the Integral Test.

Let \(f(x)=\dfrac{1}{x(\ln x)^3}\). Then \(f(n)=\dfrac{1}{n(\ln n)^3}=a_n\). Further, \(f(x)\) is positive (for \(x>1\)), continuous (because it is differentiable for \(x>0\)) and decreasing (for \(x>1\)) since \[ f'(x)=\dfrac{-[(\ln x)^3+3x(\ln x)^2]}{x^2{(\ln x)^6}} \] is negative for \(x>1\). Therefore, the Integral Test applies. We compute the integral: \[\begin{aligned} \int_2^\infty &\dfrac{1}{x(\ln x)^3}\,dx =\left. \dfrac{-1}{2(\ln x)^2}\right|_2^\infty \\ &=\lim_{b\to\infty}\left[\dfrac{-1}{2(\ln b)^2}\right]-\left[\dfrac{-1}{2(\ln 2)^2}\right] \\ &=\dfrac{1}{2(\ln 2)^2} \end{aligned}\] So \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n(\ln n)^3}\) is convergent by the Integral Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+n^2}}\)

Try a comparison.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+n^2}}\) diverges by the Limit Comparison Test with \(b_n=\dfrac{1}{n}\).

Let \(b_n=\dfrac{1}{n}\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty} \dfrac{\dfrac{1}{\sqrt[\scriptstyle 3]{n^3+n^2}}}{\dfrac{1}{n}} \\ &=\lim_{n\to\infty}\dfrac{n}{\sqrt[\scriptstyle 3]{n^3+n^2}}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\) is the divergent harmonic series, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle 3]{n^3+n^2}}\) diverges by the Limit Comparison Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(2n-1)}\)

The convergence of this series can be proven using multiple tests:
1. The Limit Comparison Test
2. The Integral Test
Try solving both ways to confirm that your answer is correct.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(2n-1)}\) converges by either of the tests:
1. The Limit Comparison Test comparing to \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n^2}\).
2. The Integral Test using partial fractions.

Using the Limit Comparison Test:
We compare to \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n^2}\) which is a convergent \(p\)-series since \(p=2>1\). Then: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{\dfrac{1}{n(2n-1)}}{\dfrac{1}{2n^2}} =\lim_{n\to\infty}\dfrac{2n^2}{n(2n-1)}=1 \end{aligned}\] Therefore, since \(0 \lt 1 \lt \infty\), \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(2n-1)}\) converges by the Limit Comparison Test.

Using the Integral Test: \[\begin{aligned} \int_1^\infty &\dfrac{1}{x(2x-1)}\,dx =\int_1^\infty \dfrac{-1}{x}+\dfrac{2}{2x-1}\,dx \\ &=\left[-\ln(x)+\ln(2x-1)\rule{0pt}{10pt}\right]_1^\infty =\left[\ln\left(\dfrac{2x-1}{x}\right)\right]_1^\infty \\ &=\ln(2)-\ln\left(\dfrac{1}{1}\right) =\ln(2) \end{aligned}\] Since the integral is finite, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(2n-1)}\) converges.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(\sqrt[\scriptstyle 3]{n})}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(\sqrt[\scriptstyle 3]{n})}\) is a convergent \(p\)-series.

We rewrite the series as: \[ \sum_{n=1}^\infty \dfrac{1}{n(\sqrt[\scriptstyle 3]{n})} =\sum_{n=1}^\infty \dfrac{1}{n^{4/3}} \] This is a \(p\)-series with \(p=\dfrac{4}{3} \gt 1\). So it converges.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle n]{n}}\)

Remember that \(\displaystyle \lim_{n\to\infty}\sqrt[\scriptstyle n]{n}=1\). What does this say about the limit of the terms?

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle n]{n}}\) diverges by the \(n^\text{th}\) Term Divergence Test.

We compute the limit of the \(n^\text{th}\) term: \[\lim_{n\to\infty}\dfrac{1}{\sqrt[\scriptstyle n]{n}}=1 \] Since this is non-zero, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptstyle n]{n}}\) diverges by the \(n^\text{th}\) Term Divergence Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^4}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^4}\) converges by the Simple Comparison Test comparing to \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^3}\).

Since \(\dfrac{\ln n}{n^4} \lt \dfrac{n}{n^4}=\dfrac{1}{n^3}\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^3}\) is a convergent \(p\)-series since \(p=3 \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{\ln n}{n^4}\) also converges by the Simple Comparison Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{n^2+2^n}{n+3^n}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{n^2+2^n}{n+3^n}\) converges by the Limit Comparison Test comparing to \(\displaystyle \sum_{n=1}^\infty \left(\dfrac{2}{3}\right)^n\).

To find a comparison series, we drop the \(n^2\) in the numerator and the \(n\) in the denominator. So we compare to \(\displaystyle \sum_{n=1}^\infty \dfrac{2^n}{3^n}\) which is a convergent geometric series since \(r=\dfrac{2}{3} \lt 1\). We compute: \[\begin{aligned} \lim_{n\to\infty}\dfrac{a_n}{b_n} &=\lim_{n\to\infty}\dfrac{\dfrac{n^2+2^n}{n+3^n}}{\dfrac{2^n}{3^n}} =\lim_{n\to\infty}\dfrac{n^2+2^n}{n+3^n}\cdot\dfrac{2^{-n}}{3^{-n}}\\ &=\lim_{n\to\infty} \dfrac{n^2 2^{-n}+1}{n 3^{-n}+1}=1 \end{aligned}\] since \(n^2 2^{-n}\) and \(n 3^{-n}\) go to \(0\). Therefore, since \(0 \lt 1 \lt \infty\), \(\displaystyle \sum_{n=1}^\infty \dfrac{n^2+2^n}{n+3^n}\) converges by the Limit Comparison Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{(n+1)!}{10^n}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{(n+1)!}{10^n}\) diverges by the Ratio Test.

We have \(a_n=\dfrac{(n+1)!}{10^n}\) and \(a_{n+1}=\dfrac{(n+2)!}{10^{n+1}}\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}\\ &=\lim_{n\to\infty}\dfrac{(n+2)!}{10^{n+1}}\cdot\dfrac{10^n}{(n+1)!}\\ &=\lim_{n\to\infty}\dfrac{n+2}{10} =\infty \end{aligned}\] Since \(\rho=\infty \gt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{(n+1)!}{10^n}\) diverges.

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\(\displaystyle \sum_{n=2}^\infty \dfrac{(\ln n)^2}{n}\)

\(\displaystyle \sum_{n=2}^\infty \dfrac{(\ln n)^2}{n}\) diverges by the Integral Test or by the Simple Comparison Test with \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\).

Using the Integral Test: The function \(f(x)=\dfrac{(\ln x)^2}{x}\) is positive, continuous and decreasing and \[ \int_2^\infty \dfrac{(\ln x)^2}{x}\,dx =\left.\dfrac{(\ln x)^3}{3}\right|_2^\infty =\infty \] Since the integral is infinite, the series \(\displaystyle \sum_{n=2}^\infty \dfrac{(\ln n)^2}{n}\) diverges.

Using the Simple Comparison Test: Since \(\dfrac{(\ln n)^2}{n}>\dfrac{1}{n}\) and \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\) is the divergent harmonic series, the series \(\displaystyle \sum_{n=2}^\infty \dfrac{(\ln n)^2}{n}\) also diverges.

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\(\displaystyle \sum_{n=0}^\infty \dfrac{1+5^n}{3^n}\)

\(\displaystyle \sum_{n=0}^\infty \dfrac{1+5^n}{3^n}\) diverges by the Simple Comparison Test, comparing to \(\displaystyle \sum_{n=0}^\infty \left(\dfrac{5}{3}\right)^n\).

Since \(\dfrac{1+5^n}{3^n} \gt \left(\dfrac{5}{3}\right)^n\) and \(\displaystyle \sum_{n=0}^\infty \left(\dfrac{5}{3}\right)^n\) is a divergent geometric series (\(r=5/3 \gt 1\)), the series \(\displaystyle \sum_{n=0}^\infty \dfrac{1+5^n}{3^n}\) diverges by the Simple Comparison Test.

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\(\displaystyle \sum_{n=2}^\infty \left(\dfrac{n-1}{n}\right)^n\)

Although \(\displaystyle \sum_{n=2}^\infty \left(\dfrac{n-1}{n}\right)^n\) has an \(n^\text{th}\) power in it, the Root Test does not prove convergence or divergence of this series since: \[ \lim_{n\to\infty}\dfrac{n-1}{n}=1 \] Therefore, you need a different test. However, we just computed \(\displaystyle \lim_{n\to\infty}\dfrac{n-1}{n}\). So try computing \(\displaystyle \lim_{n\to\infty}\left(\dfrac{n-1}{n}\right)^n\).

\(\displaystyle \sum_{n=2}^\infty \left(\dfrac{n-1}{n}\right)^n\) diverges by the \(n^\text{th}\) Term Divergence Test.

Since \(\displaystyle \lim_{n\to\infty}\dfrac{n-1}{n}=1\), we try the \(n^\text{th}\) Term Divergence Test. To compute \[ L=\lim_{n\to\infty} a_n =\lim_{n\to\infty} \left(\dfrac{n-1}{n}\right)^n \] we take the log of the limit and later exponentiate the result: \[\begin{aligned} \ln L &=\lim_{n\to\infty} n\ln\left(\dfrac{n-1}{n}\right) =\lim_{n\to\infty} \dfrac{\;\ln\left(1-\dfrac{1}{n}\right)\;}{\dfrac{1}{n}} \\ &\overset{\text{l'H}}{=}\,\lim_{n\to\infty} \dfrac{\;\dfrac{1}{1-\frac{1}{n}}\left(\dfrac{1}{n^2}\right)\;}{\dfrac{-1}{n^2}} =\lim_{n\to\infty} \dfrac{-1}{1-\dfrac{1}{n}} =-1 \end{aligned}\] Consequently, \[ L=e^{-1}\ne0 \] Therefore, \(\displaystyle \sum_{n=2}^\infty \left(\dfrac{n-1}{n}\right)^n\) diverges by the \(n^\text{th}\) Term Divergence Test.

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\(\displaystyle \dfrac{1}{3}+\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{4}{9}+\dfrac{5}{11}+...\)

The series diverges by the \(n\)th Term Test.

The series can be written in summation notation as: \[\sum_{n=1}^\infty \dfrac{n}{2n+1}\] We compute the limit of the \(n^\text{th}\) term: \[ \lim_{n\to\infty}\dfrac{n}{2n+1}=\dfrac{1}{2} \] Since \(\dfrac{1}{2}\neq0\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{2n+1}\) diverges by the \(n^\text{th}\) Term Divergence Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{(n!)^2}{(3n)!}\)

\(\displaystyle \sum_{n=1}^\infty \dfrac{(n!)^2}{(3n)!}\) converges by the Ratio Test.

We have \(a_n=\dfrac{n!n!}{(3n)!}\) and \(a_{n+1}=\dfrac{(n+1)!(n+1)!}{(3n+3)!}\). \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}= \lim_{n\to\infty}\dfrac{(n+1)!(n+1)!}{(3n+3)!}\cdot\dfrac{(3n)!}{n!n!}\\ &=\lim_{n\to\infty}\dfrac{(n+1)(n+1)}{(3n+3)(3n+2)(3n+1)} \\ &=\lim_{n\to\infty}\dfrac{n^2+2n+1}{27n^3+54x^2+33n+6}=0 \end{aligned}\] Since \(\rho=0< 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{(n!)^2}{(3n)!}\) converges by the Ratio Test.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{n+\ln n}{n^2+1}\)

The series diverges by the Limit Comparison Test, comparing with \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}\).

Keeping the largest term in the numerator and denominator, we compare to: \[ \sum_{n=1}^\infty \dfrac{n}{n^2}=\sum_{n=1}^\infty \dfrac{1}{n} \] which is the divergent harmonic series. We cannot use the Simple Comparison Test because the \(+1\) in the denominator prevents the required inequality. So we use the Limit Comparison Test: \[\begin{aligned} L&=\lim_{n\to\infty} \dfrac{a_n}{b_n} =\lim_{n\to\infty} \dfrac{(n+\ln n)}{(n^2+1)}\dfrac{n}{1} \\ &=\lim_{n\to\infty} \dfrac{n^2+n\ln n}{n^2+1}=1 \end{aligned}\] Since \(L=1\), the two series behave the same and so \(\displaystyle \sum_{n=1}^\infty \dfrac{n+\ln n}{n^2+1}\) also diverges.

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\(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{n\sqrt{n}}\)

Is there an upper bound for \(\arctan(n)\)?

\(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{n\sqrt{n}}\) converges by the Simple Comparison Test, comparing with \(\displaystyle \sum_{n=1}^\infty \dfrac{\pi}{2}\left(\dfrac{1}{n^{3/2}}\right)\)

Since \(\dfrac{\arctan n}{n\sqrt{n}} \lt \dfrac{\pi}{2}\left(\dfrac{1}{n^{3/2}}\right)\) and \(\displaystyle \sum_{n=1}^\infty \dfrac{\pi}{2}\left(\dfrac{1}{n^{3/2}}\right)\) is a convergent \(p\)-series (\(p=\dfrac{3}{2} \gt 1\)), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{\arctan n}{n\sqrt{n}}\) also converges by the Simple Comparison Test.

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\(\displaystyle \sum_{n=1}^\infty a_n\) where \(a_1=2\) and \(a_{n+1}=\dfrac{n+1}{n}a_n\)

The series diverges by the \(n^\text{th}\)-Term Divergence Test.

We write out the first \(5\) terms and generalize to the \(n^\text{th}\) term: \[\begin{aligned} a_1&=2 \quad \text{and} \quad &a_{n+1}&=\dfrac{n+1}{n}a_n \\ a_2&=\dfrac{2}{1}\cdot2 &&=2\cdot2 \\ a_3&=\dfrac{3}{2}\cdot\dfrac{2}{1}\cdot2 &&=3\cdot2 \\ a_4&=\dfrac{4}{3}\cdot\dfrac{3}{2}\cdot\dfrac{2}{1}\cdot2 &&=4\cdot2 \\ a_5&=\dfrac{5}{4}\cdot\dfrac{4}{3}\cdot\dfrac{3}{2}\cdot\dfrac{2}{1}\cdot2 &&=5\cdot2 \\ &\vdots \\ a_n&=\dfrac{n!}{(n-1)!}\cdot2=2n \end{aligned}\] But then: \[\lim_{n\to\infty}a_n=\lim_{n\to\infty}2n=\infty\] So the series diverges by the \(n^\text{th}\)-Term Divergence Test.

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