20. Convergence of Positive Series

e.1. The Ratio Test

The next test mimics the behavior of the geometric series.

Suppose \(\displaystyle \sum_{n=n_o}^\infty a_n\) is a positive series and the limit of the ratio of successive terms is \(\displaystyle \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}\).

If \(\rho<1\), then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is convergent.
If \(\rho>1\), then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is divergent.

\(\Longleftarrow\) Read this! It's easy.
\(\Longleftarrow\) The proof requires the precise definition of the limit of a sequence.

If \(\rho=1\) the Ratio Test FAILS and says nothing about \(\displaystyle \sum_{n=n_o}^\infty a_n\).

For large \(n\), \(\dfrac{a_{n+1}}{a_n}\) is approximately \(\displaystyle \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}\). So pick some large \(N\). Then for \(n \ge N\), we have \(a_n\approx a_N \rho^{n-N}\). So we can approximate the tail of the series by \[ \sum_{n=N}^\infty a_n \approx\sum_{n=N}^\infty a_N \rho^{n-N} \] This is a geometric series which converges to \[ \sum_{n=N}^\infty a_N \rho^{n-N} =\dfrac{a_N}{1-\rho} \] provided \(|\rho| \lt 1\) and diverges if \(|\rho| \ge 1\). So it is not surprizing that the original series also convegres or diverges when \(\rho\) is less than or greater than \(1\) with the exception that the borderline case of \(\rho=1\) is indeterminate instead of divergent.

Notice that the conclusions of the Ratio Test are the same as for a geometric series \(\displaystyle \sum_{n=n_o}^\infty a\rho^n\) whose ratio is \(\rho\), except that the borderline case of \(\rho=1\) is indeterminate instead of divergent.

Determine if \(\displaystyle \sum_{n=0}^\infty \dfrac{2^n}{n!}\) is convergent or divergent.

We have \(a_n=\dfrac{2^n}{n!}\) and \(a_{n+1}=\dfrac{2^{n+1}}{(n+1)!}\). So: (Notice how we divide by \(a_n\) by multiplying by its reciprocal.) \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{2^{n+1}}{(n+1)!}\cdot\dfrac{n!}{2^n} \\ &=\lim_{n\to\infty}\dfrac{2}{n+1}=0 \end{aligned}\] Since \(\rho=0 \lt 1\), the series \(\displaystyle \sum_{n=0}^\infty \dfrac{2^n}{n!}\) converges.

\(\dfrac{n!}{(n+1)!}=\dfrac{n\cdots1}{(n+1)n\cdots1}=\dfrac{1}{n+1}\)

Determine if \(\displaystyle \sum_{n=0}^\infty \dfrac{3^{2n+1}}{2n+1}\) is convergent or divergent.

We have \(a_n=\dfrac{3^{2n+1}}{2n+1}\) and \(a_{n+1}=\dfrac{3^{2(n+1)+1}}{2(n+1)+1}=\dfrac{3^{2n+3}}{2n+3}\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{3^{2n+3}}{2n+3}\cdot\dfrac{2n+1}{3^{2n+1}} \\ &=\lim_{n\to\infty}\dfrac{3^2(2n+1)}{2n+3}=9 \end{aligned}\] Since \(\rho=9>1\), the series \(\displaystyle \sum_{n=0}^\infty \dfrac{3^{2n+1}}{2n+1}\) diverges.

Determine if \(\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1}\) is convergent or divergent.

We have \(a_n=\dfrac{n}{2n^2+1}\) and \(a_{n+1}=\dfrac{n+1}{2(n+1)^2+1}\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{n+1}{2(n+1)^2+1}\cdot\dfrac{2n^2+1}{n} \\ &=\lim_{n\to\infty}\dfrac{2n^2+1}{2(n+1)^2+1}\cdot \lim_{n\to\infty}\dfrac{n+1}{n}=1\cdot1=1 \end{aligned}\] Since \(\rho=1\), the Ratio Test fails and we need to use some other test to determine the convergence.

One way is to use the Integral Test. The function \(f(n)=\dfrac{n}{2n^2+1}\) is continuous, positive and decreasing and \[ \int_0^\infty \dfrac{n}{2n^2+1}\,dn =\left. \dfrac{1}{4}\ln(2n^2+1)\right|_0^\infty=\infty \] So \(\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1}\) is divergent.

Another way is to use the Limit Comparison Test. Since \(a_n=\dfrac{n}{2n^2+1}\approx\dfrac{1}{2n}\), we compare to \(\displaystyle \sum_{n=0}^\infty b_n=\sum_{n=0}^\infty \dfrac{1}{n}\) which is the harmonic series which is divergent. So we compute the limit: \[ \lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{n}{2n^2+1}\dfrac{n}{1}=\dfrac{1}{2} \] Since \(0 \lt \dfrac{1}{2} \lt \infty\) we know \(\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1}\) is also divergent.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{n}{3^n}\) is convergent or divergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{n}{3^n}\) is convergent.

We have \(a_n=\dfrac{n}{3^n}\) and \(a_{n+1}=\dfrac{n+1}{3^{n+1}}\). So \[\begin{aligned} \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{n+1}{3^{n+1}}\cdot\dfrac{3^n}{n} \\ =\lim_{n\to\infty}\dfrac{n+1}{3n}=\dfrac{1}{3} \end{aligned}\] Since \(\rho=\dfrac{1}{3}< 1\), the series \(\displaystyle \sum_{n=0}^\infty \dfrac{n}{3^n}\) converges.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{2^nn!}{(n+1)!}\) is convergent or divergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{2^nn!}{(n+1)!}\) is divergent.

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

Notice that the Divergence Test will also work, since \[ \lim_{n\to\infty} \dfrac{2^nn!}{(n+1)!} =\lim_{n\to\infty} \dfrac{2^n}{n+1}=\infty \ne 0 \]

20. Convergence of Positive Series

e.1. The Ratio Test

Heading