Comparison Tests: Learn It 2

Giselle Miller

Comparison Tests: Learn It 2

Limit Comparison Test

The comparison test works well when we can find a suitable comparison series. However, sometimes finding an appropriate series that satisfies the required inequalities can be challenging.

Consider the series [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}[/latex]. It is natural to compare this series with the convergent series [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}[/latex].

However, we can’t use the regular comparison test because:

[latex]\frac{1}{{n}^{2}-1}>\frac{1}{{n}^{2}}[/latex]

for all integers [latex]n \geq 2[/latex]. Since our series has larger terms than the convergent comparison series, the comparison test doesn’t give us useful information.

We could search for a different comparison series, but there’s a better approach: the limit comparison test. This test allows us to compare [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}[/latex] and [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}[/latex] even when the inequality goes the “wrong” way.

The limit comparison test is particularly useful when:

The terms of your series and a known series have similar behavior for large [latex]n[/latex]
The regular comparison test fails because the inequality goes the wrong direction
Both series have the same “dominant” behavior in their numerator and denominator

The limit comparison test examines how two series relate to each other asymptotically. For two series [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] and [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] with positive terms, we evaluate: [latex]\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}[/latex].

Case 1: Finite, Non-zero Limit

If [latex]\underset{n\to \infty}{\text{lim}} \frac{a_n}{b_n} = L \neq 0[/latex], then for sufficiently large [latex]n[/latex], we have [latex]a_n \approx Lb_n[/latex]. This means the series behave similarly—either both converge or both diverge.

Let’s apply this to our example. For [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2-1}[/latex] and [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}[/latex]:

[latex]\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[/latex].

Since [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}[/latex] converges, we conclude that [latex]\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2-1}[/latex] converges.

Case 2: Limit is Zero

If [latex]\underset{n\to \infty}{\text{lim}} \frac{a_n}{b_n} = 0[/latex], then [latex]{\frac{a_n}{b_n}}[/latex] is bounded. This means there exists a constant [latex]M[/latex] such that [latex]a_n \leq Mb_n[/latex].

Therefore, if [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] converges, then [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] converges by the comparison test.

Case 3: Limit is Infinite

If [latex]\underset{n\to \infty}{\text{lim}} \frac{a_n}{b_n} = \infty[/latex], then [latex]{{a_n}{b_n}}[/latex] is unbounded. For every constant [latex]M[/latex], there exists an integer [latex]N[/latex] such that [latex]a_n \geq Mb_n[/latex] for all [latex]n \geq N[/latex].

Therefore, if [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] diverges, then [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] also diverges.

limit comparison test

Let [latex]{a}_{n},{b}_{n}\ge 0[/latex] for all [latex]n\ge 1[/latex].

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

Understanding what each limit case means can help you apply the test more effectively:

Finite limit: The series are “the same size” asymptotically
Limit is 0: The [latex]a_n[/latex] series shrinks much faster than [latex]b_n[/latex]
Limit is ∞: The [latex]a_n[/latex] series grows much faster than [latex]b_n[/latex]

When the Test Is Inconclusive

The limit comparison test doesn’t always provide information. It gives no conclusion when:

[latex]\frac{a_n}{b_n} \to 0[/latex] and [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] diverges

[latex]\frac{a_n}{b_n} \to \infty[/latex] and [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] converges

Consider the [latex]p[/latex]-series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}[/latex] and [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}[/latex].

We know that:

[latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}[/latex] diverges (since [latex]p = \frac{1}{2} \leq 1[/latex])

[latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}[/latex] converges (since [latex]p = 2 > 1[/latex])

Now suppose we try to use the convergent [latex]p[/latex]-series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}[/latex] as our comparison series:

For the divergent series: [latex]\frac{\frac{1}{\sqrt{n}}}{\frac{1}{n^3}} = \frac{n^3}{\sqrt{n}} = n^{5/2} \to \infty[/latex] as [latex]n \to \infty[/latex]
For the convergent series: [latex]\frac{\frac{1}{n^2}}{\frac{1}{n^3}} = n \to \infty[/latex] as [latex]n \to \infty[/latex]

In both cases, we get [latex]\frac{a_n}{b_n} \to \infty[/latex] with a convergent comparison series, so the limit comparison test provides no information about either series.

When the limit comparison test is inconclusive, try a different comparison series. For [latex]\frac{1}{\sqrt{n}}[/latex], compare with [latex]\frac{1}{n}[/latex] instead of [latex]\frac{1}{n^3}[/latex] to get a useful result.

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.

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

Compare this series to [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}[/latex]. Calculate

[latex]\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[/latex].

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

[latex]\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[/latex].

Therefore,

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

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

[latex]{\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}[/latex]

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

[latex]\underset{x\to \infty }{\text{lim}}\frac{\text{ln}x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0[/latex].

Therefore, [latex]\underset{n\to \infty }{\text{lim}}\text{ln}\frac{n}{n}=0[/latex], and, consequently,

[latex]\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\frac{n}{{n}^{2}}}{\frac{1}{n}}=0[/latex].

Since the limit is [latex]0[/latex] but [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] diverges, the limit comparison test does not provide any information.

Compare with [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] instead. In this case,

[latex]\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[/latex].

Since the limit is [latex]\infty[/latex] but [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] converges, the test still does not provide any information.

So now we try a series between the two we already tried. Choosing the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{3}{2}}}[/latex], we see that

[latex]\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}}[/latex].

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

[latex]\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[/latex].

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