20. Convergence of Positive Series
d.1. The Limit Comparison Test
Sometimes in applying the Simple Comparison Test to a series \(\displaystyle \sum_{n=n_o}^\infty a_n\) we may be able to identify a comparison series \(\displaystyle \sum_{n=n_o}^\infty b_n\) and show that the comparison series converges or diverges. However, we may not be able to show \(b_n \ge a_n\) or \(b_n \le a_n\) (respectively) or it may not even be true. In that case, we should try the Limit Comparison Test:
Suppose \(\displaystyle \sum_{n=n_o}^\infty a_n\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) are positive series and \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=L\).
- 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.
- 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.
- 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.
Here, “if and only if” means
if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent
then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is convergent,
and if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is divergent
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.
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.
The most common case is Case 1 where \(0 \lt L \lt \infty\). In that case, the limit \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=L\) says, for large \(n\), that \(a_n\approx L\,b_n\). So for large \(N\), the tails of the series satisfy \(\displaystyle \sum_{n=N}^\infty a_n \approx L \sum_{n=N}^\infty b_n\). So \(\displaystyle \sum_{n=n_o}^\infty a_n\) is finite if and only if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is finite.
In Case 2, the extreme case with \(L=0\), the limit \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=0\) says, for large \(n\), that \(a_n\) is much smaller than \(b_n\). So if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is finite, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is even more finite.
In Case 3, the extreme case with \(L=\infty\), the limit \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=\infty\) says, for large \(n\), that \(a_n\) is much larger than \(b_n\). So if \(\displaystyle \sum_{n=n_o}^\infty b_n\) is infinite, then \(\displaystyle \sum_{n=n_o}^\infty a_n\) is even more infinite.
Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{n^2-n-1}{n^{5/2}}\) is convergent or divergent.
This is similar to the exercise on the previous page except for the signs in the numerator. We have \(a_n=\dfrac{n^2-n-1}{n^{5/2}}\). For large \(n\), the term \(n^2\) is larger than both \(n\) and \(1\). So we take \(b_n=\dfrac{n^2}{n^{5/2}}=\dfrac{1}{n^{1/2}}\). Now \(\displaystyle \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \dfrac{1}{n^{1/2}}\) diverges because it is a \(p\)-series with \(p=\dfrac{1}{2} \lt 1\). Finally, we need to compare \(a_n\) and \(b_n\). Since, \(n^2-n-1 \lt n^2\), we have \(a_n=\dfrac{n^2-n-1}{n^{5/2}} \lt \dfrac{n^2}{n^{5/2}}=b_n\). Unfortunately, this inequality is in the wrong direction to be able to conclude \(\displaystyle \sum_{n=1}^\infty a_n\) is divergent using the Simple Comparison Test. So we try the Limit Comparison Test by computing: (Notice how we divide by \(b_n\) by multiplying by its reciprocal.) \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{n^2-n-1}{n^{5/2}}\cdot\dfrac{n^{1/2}}{1} \\ &=\lim_{n\to\infty}\dfrac{n^2-n-1}{n^2} =\lim_{n\to\infty}\left(1-\dfrac{1}{n}-\dfrac{1}{n^2}\right)=1 \end{aligned}\] Since \(0 \lt L \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty b_n\) diverges, we conclude \(\displaystyle \sum_{n=1}^\infty a_n\) also diverges.
Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{n^2+n}{n^4+\sin n}\) is convergent or divergent.
We have \(a_n=\dfrac{n^2+n}{n^4+\sin n}\). For large \(n\), the term \(n^2\) is larger than \(n\) and the term \(n^4\) is larger than \(\sin n\). So we take \(b_n=\dfrac{n^2}{n^4}=\dfrac{1}{n^2}\). Now \(\displaystyle \sum_{n=1}^\infty b_n =\sum_{n=1}^\infty \dfrac{1}{n^2}\) converges because it is a \(p\)-series with \(p=2 \gt 1\). Finally, we need to compare \(a_n\) and \(b_n\). We know \(n^2+n \gt n^2\), but we cannot say whether \(n^4+\sin n\) is bigger or smaller than \(n^4\). So we are unable to apply the Simple Comparison Test. So we try the Limit Comparison Test by computing: \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{n^2+n}{n^4+\sin n}\cdot\dfrac{n^2}{1} \\ &=\lim_{n\to\infty}\dfrac{n^4+n^3}{n^4+\sin n} =1 \end{aligned}\] Since \(0 \lt L \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty b_n\) converges, we conclude \(\displaystyle \sum_{n=1}^\infty a_n\) also converges.
Determine if \(\displaystyle \sum_{n=0}^\infty \dfrac{n^2-n}{n^{7/3}}\) is convergent or divergent.
Let \(b_n=\dfrac{n^2}{n^{7/3}}=\dfrac{1}{n^{1/3}}\).
\(\displaystyle \sum_{n=0}^\infty \dfrac{n^2-n}{n^{7/3}}\) is divergent.
We have \(a_n=\dfrac{n^2-n}{n^{7/3}}\). For large \(n\), the \(-n\) term is negligible, so we take \(b_n=\dfrac{n^2}{n^{7/3}}=\dfrac{1}{n^{1/3}}\). The series, \(\displaystyle \sum_{n=0}^\infty \dfrac{1}{n^{1/3}}\), is a divergent \(p\)-series with \(p=1/3 \lt 1\). Finally, we need to compare \(a_n\) and \(b_n\). Since, \(n^2-n \lt n^2\), we have \(a_n=\dfrac{n^2-n}{n^{7/3}} \lt \dfrac{n^2}{n^{7/3}}=b_n\). This inequality is in the wrong direction to be able to conclude \(\displaystyle \sum_{n=1}^\infty a_n\) diverges using the Simple Comparison Test. So, we try the Limit Comparison Test, \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{n^2-n}{n^{7/3}}\cdot\dfrac{n^{1/3}}{1} \\ &=\lim_{n\to\infty}\dfrac{n^2-n}{n^2} =1 \end{aligned}\] Since \(0 \lt L \lt \infty\) and \(\displaystyle \sum_{n=1}^\infty b_n\) diverges, we conclude \(\displaystyle \sum_{n=1}^\infty a_n\) also diverges.
dd
Determine if \(\displaystyle \sum_{n=0}^\infty \dfrac{2^n+n^2}{4^n-n^4}\) is convergent or divergent.
Let \(b_n=\dfrac{2^n}{4^n}=\left(\dfrac{1}{2}\right)^n\).
\(\displaystyle \sum_{n=0}^\infty \dfrac{2^n+n^2}{4^n-n^4}\) is convergent.
We have \(a_n=\dfrac{2^n+n^2}{4^n-n^4}\). For large \(n\), \(2^n\) is much larger than \(n^2\), and \(4^n\) is much larger than \(n^4\) (Try \(n=100\)). So we take \(b_n=\dfrac{2^n}{4^n}=\left(\dfrac{1}{2}\right)^n\). The series \(\displaystyle \sum_{n=0}^\infty b_n =\sum_{n=0}^\infty \left(\dfrac{1}{2}\right)^n\) is convergent because it is a geometric series with ratio \(r=\dfrac{1}{2}< 1\). To apply the Limit Comparison Test we compute \[\begin{aligned} L&=\lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{2^n+n^2}{4^n-n^4}\cdot\dfrac{4^n}{2^n} \\ &=\lim_{n\to\infty}\dfrac{2^n+n^2}{4^n-n^4}\cdot\dfrac{2^{-n}}{4^{-n}} =\lim_{n\to\infty}\dfrac{1+\dfrac{n^2}{2^n}}{1-\dfrac{n^4}{4^n}} =1 \end{aligned}\] So the series \(\displaystyle \sum_{n=0}^\infty a_n\) also converges.
The other two cases, the extreme cases with \(L=0\) or \(L=\infty\), are less common but are covered on the next page.
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