Check the convergence of $\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$ and if it converges find the sum

Question

Check whether the series converges or not. If yes, find the sum of the series $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$

My Efforts

Observe $n^3<n^4+n^2+1$

$\therefore$ $\frac{1}{n^4+n^2+n+1}<\frac{1}{n^3}$ which further implies that $\frac{n}{n^4+n^2+n+1}<\frac{n}{n^3}=\frac{1}{n^2}$

Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we conclude by comparison test that the series converges.

I am not able to find the sum of this series? Any hints or directions will be appreciated.

Answer 1

As an alternative since

$$\frac{n}{n^4+n^2+1}\sim \frac1{n^3}$$

we can conclude that the series converges by limit comparison test with $\sum \frac1{n^3}$.

For the evaluation let consider

$$\frac{n}{n^{4}+n^{2}+1} = \frac{1}{2(n^{2}-n+1)}-\frac{1}{2(n^{2}+n+1)}$$

and use telescoping, indeed for $n=k+1$ we have $$\frac{1}{n^{2}-n+1}=\frac{1}{k^{2}+k+1}$$

and therefore

$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}=\frac12\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=$$

$$=\frac12\left(1+\sum_{n=2}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12\left(1+\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12$$

Answer 2

Hint: $$ \frac{n}{n^{4}+n^{2}+1} = \frac{1}{2(n^{2}-n+1)}-\frac{1}{2(n^{2}+n+1)} $$

Answer 3

We have

$$n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$

$$((n+1)^2-(n+1)+1)=(n^2+2n+1-n-1+1)=(n^2+n+1)$$