Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? And how? What method is required? Thanks.

$\int_1^\infty \frac{1}{x^3+x}dx$
$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}$
Therfore, $A+B=0, C=0, A=1$, then $B=-1$.
$\int \frac{1}{x^3+x}dx=\int \frac{dx}{x}-\int \frac{x dx}{x^2+1}=\ln x -\frac{1}{2}\ln (x^2+1)=\ln\frac{x}{\sqrt{x^2+1}}$
For $\int_1^\infty \frac{1}{x^3+x}dx=\lim_{x \to \infty}\ln\frac{x}{\sqrt{x^2+1}}-\ln\frac{1}{\sqrt{1^2+1}}=0-\ln\frac{1}{\sqrt{2}}<\infty.$
Integral is convergent, therefore the series too.
