# Direct Comparison Test for Convergence of an Infinite Series

## Key Questions

• By Comparison Test, we can conclude that the series
${\sum}_{n = 1}^{\infty} \frac{{9}^{n}}{3 + {10}^{n}}$ converges.

Let us look at some details.
For all $n \ge q 1$,
${9}^{n} / \left\{3 + {10}^{n}\right\} \le q {9}^{n} / {10}^{n} = {\left(\frac{9}{10}\right)}^{n}$

By Geometric Series Test,
${\sum}_{n = 1}^{\infty} {\left(\frac{9}{10}\right)}^{n}$ converges since $| r | = \frac{9}{10} < 1$

Hence, by Comparison Test,
${\sum}_{n = 1}^{\infty} \frac{{9}^{n}}{3 + {10}^{n}}$ also converges.

• If an improper integral diverges, then we probably do not want to wast time trying to find its value.

Let us assume that we already know:

${\int}_{1}^{\infty} \frac{1}{x} \mathrm{dx} = \infty$

Let us look examine this uglier improper integral.

${\int}_{1}^{\infty} \frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \mathrm{dx}$

By making the numerator smaller and the denominator bigger,

$\frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \ge \frac{3 {x}^{2}}{3 {x}^{3}} = \frac{1}{x}$

By Comparison Test, we may conclude that

${\int}_{1}^{\infty} \frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \mathrm{dx}$ diverges.

(Intuitively, if the smaller integral diverges, then the larger one has no chance to converge.)

I hope that this was helpful.

• If you are trying determine the conergence of $\sum \left\{{a}_{n}\right\}$, then you can compare with $\sum {b}_{n}$ whose convergence is known.

If $0 \le q {a}_{n} \le q {b}_{n}$ and $\sum {b}_{n}$ converges, then $\sum {a}_{n}$ also converges.
If ${a}_{n} \ge q {b}_{n} \ge q 0$ and $\sum {b}_{n}$ diverges, then $\sum {a}_{n}$ also diverges.

This test is very intuitive since all it is saying is that if the larger series comverges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.