Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent?

Question

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

Answer 1

$\int_1^\infty \frac{1}{x^3+x}dx$

$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}$ 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.

Answer 2

$1/(n^3+n) \leq 1/n^3$. So by comparison test we see that this series converges