Use the Integral Test to determine whether the series is convergent or divergent. ∑n=1∞nn2+1

gonjenjemeb

Answered question

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

Beginner2021-11-13Added 15 answers

$\sum _{n=1}^{\mathrm{\infty}}\frac{n}{{n}^{2}+1}$ $f\left(x\right)=\frac{x}{{x}^{2}+1}$ 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.
${\int}_{1}^{\mathrm{\infty}}\frac{x}{{x}^{2}+1}dx=\underset{b\to \mathrm{\infty}}{lim}\frac{x}{{x}^{2}+1}dx$ $=\underset{b\to \mathrm{\infty}}{lim}\frac{1}{2}\mathrm{ln}({x}^{2}+1){\mid}_{1}^{b}$ $=\underset{b\to \mathrm{\infty}}{lim}\frac{1}{2}(\mathrm{ln}({b}^{2}+1)-\mathrm{ln}\left(2\right))$ $=\mathrm{\infty}$
The integral diverges, so by the Integral Test, the series also diverges.
Result: divergent