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
for all integers [latex]n\ge 2[/latex]. Although we could look for a different series with which to compare [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{\left({n}^{2}-1\right)}[/latex], instead we show how we can use the limit comparison test to compare
[latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}\text{ and }\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}[/latex].
Let us examine the idea behind the limit comparison test. Consider two series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex]. with positive terms [latex]{a}_{n}\text{and}{b}_{n}[/latex] and evaluate
then, for [latex]n[/latex] sufficiently large, [latex]{a}_{n}\approx L{b}_{n}[/latex]. Therefore, either both series converge or both series diverge. For the series [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{\left({n}^{2}-1\right)}[/latex] and [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}[/latex], we see that
In this case, [latex]\left\{\frac{{a}_{n}}{{b}_{n}}\right\}[/latex] is a bounded sequence. As a result, there exists a constant [latex]M[/latex] such that [latex]{a}_{n}\le M{b}_{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. On the other hand, suppose
In this case,[latex]\left\{\frac{{a}_{n}}{{b}_{n}}\right\}[/latex] is an unbounded sequence. Therefore, for every constant [latex]M[/latex] there exists an integer [latex]N[/latex] such that [latex]{a}_{n}\ge M{b}_{n}[/latex] for all [latex]n\ge N[/latex]. Therefore, if [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] diverges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges as well.
Theorem: 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.
Note that if [latex]\frac{{a}_{n}}{{b}_{n}}\to 0[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] diverges, the limit comparison test gives no information. Similarly, if [latex]\frac{{a}_{n}}{{b}_{n}}\to \infty [/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] converges, the test also provides no information. For example, consider the two series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex]. These series are both p-series with [latex]p=\frac{1}{2}[/latex] and [latex]p=2[/latex], respectively. Since [latex]p=\frac{1}{2}>1[/latex], the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}[/latex] diverges. On the other hand, since [latex]p=2<1[/latex], the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] converges. However, suppose we attempted to apply the limit comparison test, using the convergent [latex]p-\text{series}[/latex] [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}[/latex] as our comparison series. First, we see that
Therefore, if [latex]\frac{{a}_{n}}{{b}_{n}}\to \infty [/latex] when [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] converges, we do not gain any information on the convergence or divergence of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex].
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.
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
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<n[/latex], compare with [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex]. We see that
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
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,
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
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,
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.
try it
Use the limit comparison test to determine whether the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{5}^{n}}{{3}^{n}+2}[/latex] converges or diverges.
Hint
Compare with a geometric series.
Show Solution
The series diverges.
Watch the following video to see the worked solution to the above Try It.
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