20. Convergence of Positive Series

f.1. The Root Test

This test gives a different way to compare a series to a geometric series.

Suppose \(\displaystyle \sum_{n=n_o}^\infty a_n\) is a positive series and the limit of the \(n^\text{th}\) root of the \(n^\text{th}\) term is \(\displaystyle \rho=\lim_{n\to\infty}\sqrt[\small n]{a_n}\).

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 Root Test FAILS and says nothing about \(\displaystyle \sum_{n=n_o}^\infty a_n\).

For large \(n\), \(\sqrt[\small n]{a_n}\) is approximately \(\displaystyle \rho=\lim_{n\to\infty}\sqrt[\small n]{a_n}\). So pick some large \(N\). Then for \(n \ge N\), we have \(a_n\approx \rho^n\). So we can approximate the tail of the series by \[ \sum_{n=N}^\infty a_n \approx\sum_{n=N}^\infty \rho^n \] This is a geometric series which converges to \[ \sum_{n=N}^\infty \rho^n =\dfrac{\rho^N}{1-\rho} \] 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 Root Test are the same as for the Ratio Test, except \(\rho\) is defined differently.

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

We have \(a_n=\left(\dfrac{2n+1}{n+2}\right)^n\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\small n]{a_n} =\lim_{n\to\infty}\left[\left(\dfrac{2n+1}{n+2}\right)^n\right]^{1/n} \\ &=\lim_{n\to\infty}\dfrac{2n+1}{n+2}=2 \end{aligned}\] Since \(\rho=2 \gt 1\), the series \(\displaystyle \sum_{n=0}^\infty \left(\dfrac{2n+1}{n+2}\right)^n\) diverges.

Determine if \(\displaystyle \sum_{n=1}^\infty \left[\ln\left(\dfrac{2n+1}{n+2}\right)\right]^n\) is convergent or divergent.

\(\displaystyle \sum_{n=1}^\infty \left[\ln\left(\dfrac{2n+1}{n+2}\right)\right]^n\) is convergent.

We have \(a_n=\left[\ln\left(\dfrac{2n+1}{n+2}\right)\right]^n\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\small n]{a_n} =\lim_{n\to\infty}\left(\left[\ln\left(\dfrac{2n+1}{n+2}\right)\right]^n\right)^{1/n} \\ &=\lim_{n\to\infty}\ln\left(\dfrac{2n+1}{n+2}\right)=\ln2 \end{aligned}\] Since \(\rho=\ln2 \lt 1\), the series \(\displaystyle \sum_{n=1}^\infty \left[\ln\left(\dfrac{2n+1}{n+2}\right)\right]^n\) converges.

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

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

We have \(a_n=n^ne^{-n}\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\small n]{a_n} =\lim_{n\to\infty}\left(n^ne^{-n} \right)^{1/n} \\ &=\lim_{n\to\infty}\ln\left(\dfrac{n}{e}\right)=\infty \end{aligned}\] Since \(\rho=\infty \gt 1\), the series \(\displaystyle \sum_{n=0}^\infty n^ne^{-n}\) diverges.

In applying the Root Test, the following proposition is often useful:

Let \(p(n)\) be any polynomial. Then \[ \lim_{n\to\infty}\sqrt[\small n]{p(n)}=1 \]
\(\Longleftarrow\) Read this! It's easy.

We compute the limit of the log and exponentiate the result. To compute the limit of the log, we use l'Hopital's rule: \[\begin{aligned} \ln(\rho)&=\lim_{n\to\infty}\ln\sqrt[\small n]{p(n)} =\lim_{n\to\infty}\dfrac{1}{n}\ln p(n) \\ &\overset{\text{l'H}}{=}\;\lim_{n\to\infty}\dfrac{p'(n)}{p(n)}=0 \end{aligned}\] since \(p'(n)\) is a polynomial of degree one less than \(p(n)\). So \[\begin{aligned} \ln(\rho)&=0\\ \rho&=e^0=1 \end{aligned}\]

Determine if \(\displaystyle \sum_{n=3}^\infty \dfrac{2^n}{n^{100}-2^{100}}\) is convergent or divergent.

We have \(a_n=\dfrac{2^n}{n^{100}-2^{100}}\). Notice \(n^{100}-2^{100}\) is a polynomial in \(n\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\small n]{a_n} =\lim_{n\to\infty}\sqrt[\small n]{\dfrac{2^n}{n^{100}-2^{100}}} \\ &=\dfrac{\displaystyle \lim_{n\to\infty}2} {\displaystyle \lim_{n\to\infty}\sqrt[\small n]{n^{100}-2^{100}}} =\dfrac{2}{1}=2 \end{aligned}\] Since \(\rho=2 \gt 1\), the series \(\displaystyle \sum_{n=3}^\infty \dfrac{2^n}{n^{100}-2^{100}}\) diverges.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{3n^2-4n+1}{n^n}\) is convergent or divergent.

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

We have \(a_n=\dfrac{3n^2-4n+1}{n^n}\). Notice \(3n^2-4n+1\) is a polynomial in \(n\). So \[\begin{aligned} \rho&=\lim_{n\to\infty}\sqrt[\small n]{a_n} =\lim_{n\to\infty}\sqrt[\small n]{\dfrac{3n^2-4n+1}{n^n}} \\ &=\lim_{n\to\infty}\dfrac{1}{n}\sqrt[\small n]{3n^2-4n+1} =\dfrac{1}{\infty}=0 \end{aligned}\] Since \(\rho=0 \lt 1\), the series \(\displaystyle \sum_{n=1}^\infty \dfrac{3n^2-4n+1}{n^n}\) converges.

20. Convergence of Positive Series

f.1. The Root Test

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