Evaluate : $\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$ I am attempting a calculation for an integral and encounter this series:
$$S=\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$$
but I am stuck and find it difficult to evaluate its sum.
Can someone offer suggestions?
My Attempt:
Factorising: 
$$4+8k+6k^2+2k^3 = 2(k+1)(k^2+2k+2)$$ 
$S=0.3359329927...$
$1=A(k^2+2k+2)+(Bk+C)(2k+2)$
$A=1$
 A: Observe you have
\begin{align}
4+8k+6k^2+2k^3 =2(k+1)(k+1+i)(k+1-i)
\end{align}
which means
\begin{align}
\frac{1}{4+8k+6k^2+2k^3} = \frac{1}{4i(k+1)(k+1-i)}-\frac{1}{4i(k+1).(k+1+i)}
\end{align}
Then we have
\begin{align}
\sum^\infty_{k=0}\frac{1}{4+8k+6k^2+2k^3} =&\ \frac{1}{4}\sum^\infty_{k=1}\left( \frac{-i}{k(k-i)}+\frac{i}{k(k+i)}\right)\\
=&\ \frac{1}{4}\left( \psi_0(1-i)+\psi_0(1+i)+2\gamma\right)
\end{align}
where $\psi_0$ is the digamma function and $\gamma$ is the Euler–Mascheroni constant.
I have used the fact that
\begin{align}
\psi_0(z+1) = -\gamma+\sum^\infty_{k=1}\frac{z}{k(k+z)}.
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
S & \equiv \sum_{k = 0}^{\infty}{1 \over 4 + 8k + 6k^{2} + 2k^{3}} =
{1 \over 2}\sum_{k = 1}^{\infty}{1 \over k^{3} + k} =
{1 \over 2}\Im\sum_{k = 1}^{\infty}{1 \over k\pars{k - \ic}}
\\[5mm] & =
{1 \over 2}\,\Im\sum_{k = 0}^{\infty}{1 \over \pars{k + 1}\pars{k + 1 - \ic}} =
{1 \over 2}\,\Im\bracks{\Psi\pars{1} - \Psi\pars{1 - \ic} \over \ic} =
{1 \over 2}\,\Im\bracks{\ic\gamma + \ic\Psi\pars{1 - \ic}}
\\[5mm] & = \bbx{\gamma + \Re\Psi\pars{1 - \ic} \over 2} =
\bbx{{1 \over 2}\,\Re\pars{H_{-\ic}}} \approx 0.3359
\end{align}
