3.6 Direct Comparison Test

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We will use the DCT to determine if an infinite series converges or diverges.

Direct Comparison Test

In this section, we will determine whether a given series (with positive terms) converges or diverges by comparing it to a series whose behavior is known. Thus, it is important to recall the basic facts about $p$ -series and geometric series:

$p$ -series: $\displaystyle \sum _{n=1}^\infty \frac {1}{n^p}$ , converges if $p>1$ and diverges if $p\leq 1$
geometric series: $\displaystyle \sum _{n=0}^\infty ar^n$ , converges if $|r|<1$ and diverges if $|r|\geq 1$

In the statement of the theorem below, the series $\displaystyle \sum _{n=1}^\infty a_n$ is given and we are to determine whether it converges or diverges. To that end, we will compare the given series to a series of our choosing, denoted in the theorem by $\displaystyle \sum _{n=1}^\infty b_n$ .

Direct Comparison Test Suppose $a_n, b_n \geq 0$ for all $n$ .

If $\; \displaystyle {\sum _{n=1}^\infty b_n}$ converges, and $a_n \leq b_n$ for all $n$ , then $\displaystyle {\sum _{n=1}^\infty a_n}$ converges.
In words: less than a convergent series is convergent.
If $\; \displaystyle {\sum _{n=1}^\infty b_n}$ diverges, and $a_n \geq b_n$ for all $n$ , then $\displaystyle {\sum _{n=1}^\infty a_n}$ diverges.
In words: greater than a divergent series is divergent.

Unfortunately, the Direct Comparison Test is sometimes inconclusive, as noted in the following remark.

If $\; \displaystyle {\sum _{n=1}^\infty b_n}$ converges, and $a_n \geq b_n$ for all $n$ , then no conclusion can be made regarding $\displaystyle {\sum _{n=1}^\infty a_n}$
In words: greater than a convergent series is inconclusive.
If $\; \displaystyle {\sum _{n=1}^\infty b_n}$ diverges, and $a_n \leq b_n$ for all $n$ , then no conclusion can be made regarding $\displaystyle {\sum _{n=1}^\infty a_n}$
In words: less than a divergent series is inconclusive.

The qualifying statement for all $n$ associated with the conditions $a_n \leq b_n$ and $a_n \geq b_n$ can be relaxed to for $n$ sufficiently large.

Before we look at some examples, we review some basic facts about inequalities:

(a): If $0 < a< b$ then $\frac {1}{a} > \frac {1}{b}$
(b): If $0<a<b$ and $c>0$ then $ac< bc$

example 1 Determine if the series $\sum _{n=1}^\infty \frac {1}{2n^2 + 3}$ converges or diverges.
We will use the DCT with the series $\sum _{n=1}^\infty \frac {1}{n^2}.$ This is a $p$ -series with $p=2 >1$ and so it converges. To make the comparison, first note that $2n^2 + 3 > n^2 > 0$ for all $n \geq 1$ . Taking reciprocals yields $\frac {1}{2n^2 + 3} < \frac {1}{n^2} \; \; \text { for } \; \; n \geq 1,$ and so by the DCT (case 1), $\sum _{n=1}^\infty \frac {1}{2n^2 + 3}$ also converges.

(problem 1) Determine if the series converges or diverges: $\;\displaystyle {\sum _{n=1}^\infty \frac {1}{n^3 + 1}}$

Which series should we compare this to?

$\displaystyle \sum _{n=1}^\infty \frac {1}{n^2}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{3^n}$

Which way does the comparison go?

$\displaystyle {\frac {1}{n^3 + 1} \leq \frac {1}{n^3}}$ for $n \geq 1$ $\displaystyle {\frac {1}{n^3 + 1} \geq \frac {1}{n^3}}$ for $n \geq 1$

Describe the behavior of the series $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^3 + 1}:}$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 2 Determine if the series $\sum _{n=0}^\infty \frac {3^n}{4^n + 1}$ converges or diverges.
We will use the DCT with the series $\sum _{n=0}^\infty \left (\frac 34\right )^n.$ This is a geometric series with common ratio, $r = 3/4$ . Since $-1 < \frac 34 < 1$ , this series converges. To make the comparison, first note that $4^n + 1 > 4^n > 0$ for all $n \geq 0$ . By taking the reciprocal of both sides, we have $\frac {1}{4^n + 1} < \frac {1}{4^n} \; \; \text { for } \;\; n \geq 0,$ and multiplying by $3^n$ we get $\frac {3^n}{4^n + 1} < \frac {3^n}{4^n} = \left (\frac 34\right )^n \;\; \text { for } \;\; n \geq 0.$ Therefore, by the DCT (case 1), $\sum _{n=0}^\infty \frac {3^n}{4^n + 1}$ also converges.

(problem 2) Determine if the series converges or diverges: $\;\displaystyle \sum _{n=0}^\infty \frac {1}{3^n + 1}$

Which series should we compare this to?

$\displaystyle \sum _{n=0}^\infty \frac {1}{2^n}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=0}^\infty \frac {1}{3^n}$

Which way does the comparison go?

$\frac {1}{3^n + 1} \leq \frac {1}{3^n}$ for $n \geq 0$ $\frac {1}{3^n + 1} \geq \frac {1}{3^n}$ for $n \geq 0$

Describe the behavior of the series $\sum _{n=0}^\infty \frac {1}{3^n + 1}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 3 Determine if the series $\sum _{n=1}^\infty \frac {\sin ^2(n)}{n^3}$ converges or diverges.
We will use the DCT with the series $\sum _{n=1}^\infty \frac {1}{n^3}.$ This is a $p$ -series with $p=3 >1$ and so it converges. To make the comparison, first note that $0 \leq \sin ^2(n) \leq 1$ for all $n \geq 1$ . Dividing by $n^3$ , we have $\frac {\sin ^2(n)}{n^3} \leq \frac {1}{n^3} \; \; \text { for } \; \; n \geq 1,$ and so by the DCT (case 1), $\sum _{n=1}^\infty \frac {\sin ^2(n)}{n^3}$ also converges.

(problem 3) Determine if the series converges or diverges: $\sum _{n=0}^\infty \frac {\cos ^2(n)}{2^n + 1}$
Which series should we compare this to?

$\displaystyle \sum _{n=0}^\infty \frac {1}{2^n}$ $\displaystyle \sum _{n=0}^\infty \frac {\cos ^2(n)}{2^n}$ $\displaystyle \sum _{n=0}^\infty \frac {1}{3^n}$

Which way does the comparison go?

$\frac {\cos ^2(n)}{2^n + 1}\leq \frac {1}{2^n}$ for $n \geq 0$ $\frac {\cos ^2(n)}{2^n + 1} \geq \frac {1}{2^n}$ for $n \geq 0$

Describe the behavior of the series $\sum _{n=0}^\infty {\cos ^2(n)}{2^n + 1}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 4 Determine if the series $\sum _{n=1}^\infty \frac {\cos ^4(n)}{n}$ converges or diverges.
We will use the DCT with the divergent harmonic series $\sum _{n=1}^\infty \frac {1}{n}.$

To make the comparison, first note that $0 \leq \cos ^4(n) \leq 1$ for all $n \geq 1$ . Dividing by $n$ , we have $\frac {\cos ^4(n)}{n} \leq \frac {1}{n} \; \; \text { for } \; \; n \geq 1.$ Noting that this comparison goes in the wrong direction (smaller than a divergent), the DCT gives no conclusion about the behavior of the series $\sum _{n=1}^\infty \frac {\cos ^4(n)}{n}$ .

(problem 4) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {n^2}{n^3 + 1}}$
Which series should we compare this to?

$\displaystyle \sum _{n=1}^\infty \frac {1}{n}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^3}$

Which way does the comparison go?

$\frac {n^2}{n^3 + 1} \leq \frac {1}{n}$ for $n \geq 1$ $\frac {n^2}{n^3 + 1} \geq \frac {1}{n}$ for $n \geq 1$

Describe the behavior of the series $\sum _{n=1}^\infty \frac {n^2}{n^3 + 1}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 5 Determine if the series $\sum _{n=2}^\infty \frac {1}{\sqrt {n^2 -1}}$ converges or diverges.
We will use the DCT with the divergent harmonic series $\sum _{n=2}^\infty \frac {1}{n}.$

To make the comparison, first note that $0 < \sqrt {n^2 -1} < \sqrt {n^2} = n$ for all $n \geq 2$ . By taking the reciprocal of both sides, we have $\frac {1}{\sqrt {n^2 -1}} > \frac {1}{n},$ and so by the DCT (case 2), $\sum _{n=2}^\infty \frac {1}{\sqrt {n^2 -1}}$ also diverges.

(problem 5) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {n+3}{n^2}}$
Which series should we compare this to?

$\displaystyle \sum _{n=1}^\infty \frac {1}{n}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2}$ $\displaystyle \sum _{n=1}^\infty \frac {3}{n^2}$

Which way does the comparison go?

$\frac {n+3}{n^2} \leq \frac {1}{n}$ for $n \geq 1$ $\frac {n+3}{n^2} \geq \frac {1}{n}$ for $n \geq 1$

Describe the behavior of the series $\sum _{n=1}^\infty \frac {n+3}{n^2}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 6 Determine if the series $\sum _{n=0}^\infty \frac {3^n + 5}{2^n}$ converges or diverges.
We will use the DCT with the divergent geometric series $\sum _{n=0}^\infty \left (\frac 32\right )^n.$

To make the comparison, first note that $3^n + 5 > 3^n > 0$ for all $n \geq 0$ . Dividing by $2^n$ , we have $\frac {3^n + 5}{2^n} > \frac {3^n}{2^n} = \left (\frac 32\right )^n \;\; \text { for } \;\; n \geq 0,$ and so by the DCT (case 2), $\sum _{n=0}^\infty \frac {2^n + 5}{3^n}$ also diverges.

(problem 6) Determine if the series converges or diverges: $\displaystyle {\sum _{n=0}^\infty \frac {5^n + 4}{3^n}}$
Which series should we compare this to?

$\displaystyle \sum _{n=0}^\infty \left (\frac 43\right )^n$ $\displaystyle \sum _{n=0}^\infty \left (\frac 53\right )^n$ $\displaystyle \sum _{n=0}^\infty \left (\frac 13\right )^n$

Which way does the comparison go?

$\frac {5^n + 4}{3^n} \leq \left (\frac 53\right )^n$ for $n \geq 0$ $\frac {5^n + 4}{3^n}\geq \left (\frac 53\right )^n$ for $n \geq 0$

Describe the behavior of the series $\sum _{n=0}^\infty \frac {5^n + 4}{3^n}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

example 7 Determine if the series $\sum _{n=2}^\infty \frac {1}{\ln (n)}$ converges or diverges.
We will use the DCT with the series $\sum _{n=2}^\infty \frac {1}{n\ln (n)}$ which diverges (as shown in the section on the Integral Test). To make the comparison, first note that $0 < \ln (n) < n\ln (n)$ for all $n \geq 2$ . Taking reciprocals, we have $\frac {1}{\ln (n)} > \frac {1}{n\ln (n)} \;\; \text { for } \;\; n \geq 2,$ and so by the DCT (case 2), $\sum _{n=2}^\infty \frac {1}{\ln (n)}$ also diverges.

example 8 Determine if the series $\sum _{n=1}^\infty \frac {n+2}{n^3}$ converges or diverges.
We will use the DCT with the convergent p-series $\sum _{n=1}^\infty \frac {1}{n^2}.$

To make the comparison, first note that $n+2 > n$ for all $n \geq 1$ . Dividing by $n^3$ , we have $\frac {n+2}{n^3} > \frac {n}{n^3} = \frac {1}{n^2} \; \; \text { for } \; \; n \geq 1.$ Noting that this comparison goes in the wrong direction (larger than a convergent), the DCT gives no conclusion about the behavior of the series $\sum _{n=1}^\infty \frac {n+2}{n^3}$ . In the next section on the Limit Comparison Test, we will learn a technique that allows us to make a conclusion about this series by means of a slightly different type of comparison.

(problem 8) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {n^3 +3}{n^6}}$
Which series should we compare this to?

$\displaystyle \sum _{n=1}^\infty \frac {1}{n^2}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=1}^\infty \frac {3}{n^6}$

Which way does the comparison go?

$\frac {n^3 +3}{n^6} \leq \frac {1}{n^3}$ for $n \geq 1$ $\frac {n^3+3}{n^6} \geq \frac {1}{n^3}$ for $n \geq 1$

Describe the behavior of the series $\sum _{n=1}^\infty \frac {n^3+3}{n^6}:$

Converges by DCT Diverges by DCT No Conclusion from DCT

Video Lesson

Here is a detailed, lecture style video on the Direct Comparison Test:

Direct Comparison Test

Video Lesson

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