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

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{n}{n^2+1}}$
${\displaystyle f\left(x\right)=\frac{x}{x^2+1}}$ is decrasing since the degree of the denominator is higher than the degree of the numerator. It is also positive and continuous for ${\displaystyle x\ge 0}$ so we can use the Integral Test.
${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{x}{x^2+1}dx=\underset{b\to \mathrm{\infty }}{lim}\frac{x}{x^2+1}dx}$
${\displaystyle =\underset{b\to \mathrm{\infty }}{lim}\frac{1}{2}\ln (x^2+1){\mid }_1^b}$
${\displaystyle =\underset{b\to \mathrm{\infty }}{lim}\frac{1}{2}(\ln (b^2+1)-\ln \left(2\right))}$
${\displaystyle =\mathrm{\infty }}$
The integral diverges, so by the Integral Test, the series also diverges.
Result: divergent