Find the sum $\sum_{n=1}^{\infty} \frac{4n}{n^4+2n^2+9}$ Find the sum
$$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$
By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that
$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2+n)}$ for all $n\ge n_0$, $n_0$ is large enough.
And $\sum_{n=1}^{\infty} \dfrac{5}{6(n^4+n^2)}= \sum_{n=1}^{\infty}\dfrac{5}{6(n^2+n)}=\dfrac{5}{6}$.
But it is not enough to confirm that the given series converges to $\dfrac{5}{6}$. Can someone help me, please? Thanks in advanced.

HINT:
$$(n^2)^2+3^2+2n^2=(n^2+3)^2-(2n)^2=(n^2+2n+3)(n^2-2n+3)$$
$$(n^2+2n+3)-(n^2-2n+3)=?$$
Now if $f(m)=m^2-2m+3,$
$f(m+2)=(m+2)^2-2(m+2)+3=m^2+2m+3$
