gonjenjemeb

2021-11-12

Use the Integral Test to determine whether the series is convergent or divergent.
$\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{n}^{2}+1}$

### Answer & Explanation

Florence Pittman

$\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{n}^{2}+1}$
$f\left(x\right)=\frac{x}{{x}^{2}+1}$ is decrasing since the degree of the denominator is higher than the degree of the numerator. It is also positive and continuous for $x\ge 0$ so we can use the Integral Test.
${\int }_{1}^{\mathrm{\infty }}\frac{x}{{x}^{2}+1}dx=\underset{b\to \mathrm{\infty }}{lim}\frac{x}{{x}^{2}+1}dx$
$=\underset{b\to \mathrm{\infty }}{lim}\frac{1}{2}\mathrm{ln}\left({x}^{2}+1\right){\mid }_{1}^{b}$
$=\underset{b\to \mathrm{\infty }}{lim}\frac{1}{2}\left(\mathrm{ln}\left({b}^{2}+1\right)-\mathrm{ln}\left(2\right)\right)$
$=\mathrm{\infty }$
The integral diverges, so by the Integral Test, the series also diverges.
Result: divergent

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