Question

use the integral test to determine whether the series is convergent or divergent.
evaluate the following integral.
sum _ { n = 1 } ^ { infty } \frac { n } { n ^ { 2 } + 16 }
int _ { 1 } ^ { infty } \frac { x } { x ^ { 2 } + 16 } d x
since the integral --select-- finite, the series is --select--.

Explanation:

Step1: Apply substitution

Let \(u = x^{2}+16\), then \(du=2x dx\), and \(x dx=\frac{1}{2}du\).
When \(x = 1\), \(u=1 + 16=17\); when \(x=\infty\), \(u=\infty\).
The integral \(\int_{1}^{\infty}\frac{x}{x^{2}+16}dx=\frac{1}{2}\int_{17}^{\infty}\frac{du}{u}\).

Step2: Evaluate the integral

We know that \(\int\frac{du}{u}=\ln|u|+C\).
So \(\frac{1}{2}\int_{17}^{\infty}\frac{du}{u}=\frac{1}{2}\lim_{b
ightarrow\infty}\int_{17}^{b}\frac{du}{u}\).
\(\frac{1}{2}\lim_{b
ightarrow\infty}[\ln u]_{17}^{b}=\frac{1}{2}\lim_{b
ightarrow\infty}(\ln b-\ln17)\).
Since \(\lim_{b
ightarrow\infty}\ln b=\infty\), \(\frac{1}{2}\lim_{b
ightarrow\infty}(\ln b-\ln17)=\infty\).

Answer:

Since the integral \(\int_{1}^{\infty}\frac{x}{x^{2}+16}dx\) is not finite (it diverges), the series \(\sum_{n = 1}^{\infty}\frac{n}{n^{2}+16}\) is divergent.