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:

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.

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.

[×]

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.

$\Longleftarrow$ 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.

Justification

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.

Hint

Let $b_n=\dfrac{n^2}{n^{7/3}}=\dfrac{1}{n^{1/3}}$ .

[×]

Answer

$\displaystyle \sum_{n=0}^\infty \dfrac{n^2-n}{n^{7/3}}$ is divergent.

[×]

Solution

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.

[×]

Determine if $\displaystyle \sum_{n=0}^\infty \dfrac{2^n+n^2}{4^n-n^4}$ is convergent or divergent.

Hint

Let $b_n=\dfrac{2^n}{4^n}=\left(\dfrac{1}{2}\right)^n$ .

[×]

Answer

$\displaystyle \sum_{n=0}^\infty \dfrac{2^n+n^2}{4^n-n^4}$ is convergent.

[×]

Solution

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.

20. Convergence of Positive Series

d.1. The Limit Comparison Test

Justification

Hint

Answer

Solution

Hint

Answer

Solution

Heading