20. Convergence of Positive Series

d.2. The Limit Comparison Test - Extreme Cases

Suppose \(\displaystyle \sum_{n=n_o}^\infty a_n\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) are positive series and \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=L\).

  1. If \(0 \lt L \lt \infty\), then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is convergent if and only if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent.
  2. If \(L=0\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is also convergent.
  3. If \(L=\infty\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is divergent, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is also divergent.

Cases (2) and (3) are called the extreme cases, and arise very rarely.
If \(L=0\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is divergent, or \(L=\infty\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent, the Limit Comparison Test FAILS.

We now look at the extreme cases, \(L=0\) and \(L=\infty\) which are less common.

If \(L=0\), then \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=0\) says the \(a_n\) are much smaller than the \(b_n\). So if \(\displaystyle \sum_{n=n_o}^\infty b_n\) converges, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) also converges.

Similarly, if \(L=\infty\), then \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=\infty\) says the \(a_n\) are much bigger than the \(b_n\). So if \(\displaystyle \sum_{n=n_o}^\infty b_n\) diverges, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) also diverges.

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

We try to compare to \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^2}\) which is a convergent \(p\)-series since \(p=2 \gt 1\). Thus, we take \(a_n=\dfrac{\ln n}{n^2}\) and \(b_n=\dfrac{1}{n^2}\). We compute \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{\ln n}{n^2}\cdot\dfrac{n^2}{1} \\ &=\lim_{n\to\infty}\ln n=\infty \end{aligned}\] Since \(L=\infty\) but \(\displaystyle \sum_{n=2}^\infty b_n=\sum_{n=2}^\infty \dfrac{1}{n^2}\) is convergent, the Limit Comparison Test FAILS! Back to the drawing board.

This time we compare to \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n^{3/2}}\) which is also a convergent \(p\)-series since \(p=\dfrac{3}{2}>1\).  

We have \(\begin{aligned}\dfrac{\ln n}{n^2} \gt \dfrac{1}{n^2}\end{aligned}\) but to show convergence we need \(\dfrac{\ln n}{n^2} \lt b_n\). So we make \(b_n\) slightly bigger by making the power slightly smaller (\(\begin{aligned}\dfrac{3}{2}\end{aligned}\) is slightly smaller than \(2\)) while still keeping the series convergent.

Thus, we take \(a_n=\dfrac{\ln n}{n^2}\) and \(b_n=\dfrac{1}{n^{3/2}}\). Using l'Hopital's Rule, we compute \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{\ln n}{n^2}\cdot\dfrac{n^{3/2}}{1}\\ &=\lim_{n\to\infty}\dfrac{\ln n}{n^{1/2}} \;\overset{\text{l'H}}{=}\; \lim_{n\to\infty}\dfrac{1/n}{\dfrac{1}{2}n^{-1/2}} \\ &=\lim_{n\to\infty}\dfrac{2n^{1/2}}{n} =\lim_{n\to\infty}\dfrac{2}{n^{1/2}}=0 \end{aligned}\] Since \(L=0\) and \(\displaystyle \sum_{n=2}^\infty b_n =\sum_{n=2}^\infty \dfrac{1}{n^{3/2}}\) is convergent, the Limit Comparison Test says \(\displaystyle \sum_{n=2}^\infty a_n =\sum_{n=2}^\infty \dfrac{\ln n}{n^2}\) is also convergent.

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

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

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

We compare to \(\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}\) which is a harmonic series and hence divergent. We take \(a_n=\dfrac{1}{\ln n}\) and \(b_n=\dfrac{1}{n}\). Using l'Hopital's Rule, we compute \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{1}{\ln n}\dfrac{n}{1} =\lim_{n\to\infty}\dfrac{n}{\ln n} \\ &\overset{\text{l'H}}{=}\;\lim_{n\to\infty}\dfrac{1}{1/n} =\lim_{n\to\infty}n=\infty \end{aligned}\] Since \(L=\infty\) and \(\displaystyle \sum_{n=2}^\infty b_n =\sum_{n=2}^\infty\dfrac{1}{n}\) diverges, we conclude \(\displaystyle \sum_{n=2}^\infty a_n =\sum_{n=2}^\infty \dfrac{1}{\ln n}\) also diverges.

Notice the Simple Comparison Test would also work here.

© MYMathApps

Supported in part by NSF Grant #1123255