Evaluate : $\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$

Question

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: $$4+8k+6k^2+2k^3 = 2(k+1)(k^2+2k+2)$$

$S=0.3359329927...$

$1=A(k^2+2k+2)+(Bk+C)(2k+2)$

$A=1$

Answer 1

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}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}