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.
 A: 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 $f(m)=m^2-2m+3,$
$f(m+2)=(m+2)^2-2(m+2)+3=m^2+2m+3$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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$\ds{\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9}}$.

\begin{align}
&\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9}  =
\sum_{n = 0}^{\infty}{4n \over \pars{n^{2} - 2n + 3}\pars{n^{2} + 2n + 3}}
=
\sum_{n = 0}^{\infty}\pars{{1 \over n^{2} - 2n + 3} - {1 \over n^{2} + 2n + 3}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{\sum_{n = 0}^{N}{1 \over \pars{n - 1}^{2} + 2} -
\sum_{n = 0}^{N}{1 \over \pars{n + 1}^{2} + 2}} =
\lim_{N \to \infty}\pars{\sum_{n = -1}^{N - 1}{1 \over n^{2} + 2} -
\sum_{n = 1}^{N + 1}{1 \over n^{2} + 2}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{{1 \over 3} + {1 \over 2} +
\sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} -
\sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} - {1 \over N^{2} + 2} -
{1 \over \pars{N + 1}^{2} + 2}}
\\[5mm] = &\
\lim_{N \to \infty}\pars{{5 \over 6} - {1 \over N^{2} + 2} -
{1 \over \pars{N + 1}^{2} + 2}} = \bbx{\ds{5 \over 6}}
\end{align}

