Limit Comparison Test

Learning Outcomes

Use the limit comparison test to determine convergence of a series

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

\sum_{n = 2}^{\infty} \frac{1}{n^{2} - 1}

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}$ .

It is natural to compare this series with the convergent series

\sum_{n = 2}^{\infty} \frac{1}{n^{2}}

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}$ .

However, this series does not satisfy the hypothesis necessary to use the comparison test because

\frac{1}{n^{2} - 1} > \frac{1}{n^{2}}

$\frac{1}{{n}^{2}-1}>\frac{1}{{n}^{2}}$

for all integers $n\ge 2$ . Although we could look for a different series with which to compare $\displaystyle\sum _{n=2}^{\infty }\frac{1}{\left({n}^{2}-1\right)}$ , instead we show how we can use the limit comparison test to compare

\sum_{n = 2}^{\infty} \frac{1}{n^{2} - 1} and \sum_{n = 2}^{\infty} \frac{1}{n^{2}}

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}\text{ and }\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}$ .

Let us examine the idea behind the limit comparison test. Consider two series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ . with positive terms ${a}_{n}\text{and}{b}_{n}$ and evaluate

lim_{n \to \infty} \frac{a_{n}}{b_{n}}

$\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}$ .

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L \neq 0

$\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=L\ne 0$ ,

then, for $n$ sufficiently large, ${a}_{n}\approx L{b}_{n}$ . Therefore, either both series converge or both series diverge. For the series $\displaystyle\sum _{n=2}^{\infty }\frac{1}{\left({n}^{2}-1\right)}$ and $\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}$ , we see that

lim_{n \to \infty} \frac{\frac{1}{(n^{2} - 1)}}{\frac{1}{n^{2}}} = lim_{n \to \infty} \frac{n^{2}}{n^{2} - 1} = 1

$\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{\left({n}^{2}-1\right)}}{\frac{1}{{n}^{2}}}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{2}}{{n}^{2}-1}=1$ .

Since $\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}$ converges, we conclude that

\sum_{n = 2}^{\infty} \frac{1}{n^{2} - 1}

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}$

converges.

The limit comparison test can be used in two other cases. Suppose

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = 0

$\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=0$ .

In this case, $\left\{\frac{{a}_{n}}{{b}_{n}}\right\}$ is a bounded sequence. As a result, there exists a constant $M$ such that ${a}_{n}\le M{b}_{n}$ . Therefore, if $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ converges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges. On the other hand, suppose

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \infty

$\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=\infty$ .

In this case, $\left\{\frac{{a}_{n}}{{b}_{n}}\right\}$ is an unbounded sequence. Therefore, for every constant $M$ there exists an integer $N$ such that ${a}_{n}\ge M{b}_{n}$ for all $n\ge N$ . Therefore, if $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ diverges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges as well.

Theorem: Limit Comparison Test

Let ${a}_{n},{b}_{n}\ge 0$ for all $n\ge 1$ .

If $\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=L\ne 0$ , then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ both converge or both diverge.
If $\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=0$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ converges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges.
If $\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=\infty$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ diverges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.

Note that if $\frac{{a}_{n}}{{b}_{n}}\to 0$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ diverges, the limit comparison test gives no information. Similarly, if $\frac{{a}_{n}}{{b}_{n}}\to \infty$ and $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ converges, the test also provides no information. For example, consider the two series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ and $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ . These series are both p-series with $p=\frac{1}{2}$ and $p=2$ , respectively. Since $p=\frac{1}{2}>1$ , the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges. On the other hand, since $p=2<1$ , the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p-\text{series}$ $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}$ as our comparison series. First, we see that

\frac{\frac{1}{\sqrt{n}}}{\frac{1}{n^{3}}} = \frac{n^{3}}{\sqrt{n}} = n^{\frac{5}{2}} \to \infty as n \to \infty

$\frac{\frac{1}{\sqrt{n}}}{\frac{1}{{n}^{3}}}=\frac{{n}^{3}}{\sqrt{n}}={n}^{\frac{5}{2}}\to \infty \text{as}n\to \infty$ .

Similarly, we see that

\frac{\frac{1}{n^{2}}}{\frac{1}{n^{3}}} = n \to \infty as n \to \infty

$\frac{\frac{1}{{n}^{2}}}{\frac{1}{{n}^{3}}}=n\to \infty \text{as}n\to \infty$ .

Therefore, if $\frac{{a}_{n}}{{b}_{n}}\to \infty$ when $\displaystyle\sum _{n=1}^{\infty }{b}_{n}$ converges, we do not gain any information on the convergence or divergence of $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ .

Example: Using the Limit Comparison Test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$
$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$
$\displaystyle\sum _{n=1}^{\infty }\frac{\text{ln}\left(n\right)}{{n}^{2}}$

Show Solution

Compare this series to $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ . Calculate

$\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{\left(\sqrt{n}+1\right)}}{\frac{1}{\sqrt{n}}}=\underset{n\to \infty }{\text{lim}}\begin{array}{c}\frac{\sqrt{n}}{\sqrt{n}+1}\\ \end{array}=\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}=1$ .

By the limit comparison test, since $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges, then $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$ diverges.
Compare this series to $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}$ . We see that

$\underset{n\to \infty }{\text{lim}}\frac{\frac{\left({2}^{n}+1\right)}{{3}^{n}}}{\frac{{2}^{n}}{{3}^{n}}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{3}^{n}}\cdot \frac{{3}^{n}}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\left[1+{\left(\frac{1}{2}\right)}^{n}\right]=1$ .

Therefore,

$\underset{n\to \infty }{\text{lim}}\frac{\frac{\left({2}^{n}+1\right)}{{3}^{n}}}{\frac{{2}^{n}}{{3}^{n}}}=1$ .

Since $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}$ converges, we conclude that $\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$ converges.
Since [latex]\text{ln}n

${\underset{n\to\infty}\lim} \text{ln}\frac{n}{n^2}\frac{1}{n}= {\underset{n\to\infty}\lim} \frac{\text{ln}n}{n^2}\cdot \frac{n}{1}= {\underset{n\to\infty}\lim} \frac{\text{ln}n}{n}$

In order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\frac{n}{n}$ , evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\frac{\left(x\right)}{x}$ . These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain

$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0$ .

Therefore, $\underset{n\to \infty }{\text{lim}}\text{ln}\frac{n}{n}=0$ , and, consequently,

$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\frac{n}{{n}^{2}}}{\frac{1}{n}}=0$ .

Since the limit is $0$ but $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ diverges, the limit comparison test does not provide any information.
Compare with $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ instead. In this case,

$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\frac{n}{{n}^{2}}}{\frac{1}{{n}^{2}}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}n}{{n}^{2}}\cdot \frac{{n}^{2}}{1}=\underset{n\to \infty }{\text{lim}}\text{ln}n=\infty$ .

Since the limit is $\infty$ but $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges, the test still does not provide any information.
So now we try a series between the two we already tried. Choosing the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{3}{2}}}$ , we see that

$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\frac{n}{{n}^{2}}}{\frac{1}{{n}^{\frac{3}{2}}}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}n}{{n}^{2}}\cdot \frac{{n}^{\frac{3}{2}}}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}n}{\sqrt{n}}$ .

As above, in order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\frac{n}{\sqrt{n}}$ , evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\frac{x}{\sqrt{x}}$ . Using L’Hôpital’s rule,

$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}x}{\sqrt{x}}=\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x}=\underset{x\to \infty }{\text{lim}}\frac{2}{\sqrt{x}}=0$ .

Since the limit is $0$ and $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{3}{2}}}$ converges, we can conclude that $\displaystyle\sum _{n=1}^{\infty }\frac{\text{ln}n}{{n}^{2}}$ converges.

try it

Use the limit comparison test to determine whether the series $\displaystyle\sum _{n=1}^{\infty }\frac{{5}^{n}}{{3}^{n}+2}$ converges or diverges.

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It.

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Module 5: Sequences and Series