<s>[INST] Finding a generalized form for this series While i was just playing around with series i came across this one,
$$
S = \sum_{k=1}^\infty[\frac{k}{k-\frac{1}{2}}+\frac{k-\frac{1}{2}}{k}-\frac{k+\frac{1}{2}}{k} - \frac{k}{k-\frac{1}{2}}]
$$
Which simplified nicely into
$$
S=\sum_{k=1}^{\infty}\frac{1}{4k^3-k} = \ln(4)-1
$$
Then i wondered the results would be if i changed the $\frac{1}{2}$ into $\frac{1}{n}$
I simplified the initial summation to 
$$
S_n = \sum_{k=1}^{\infty}\frac{2}{n^3 k^3 - nk}
$$
Then i found a few of the results of the summations,
$S_2 = \ln(4)-1$
$S_3 = \ln(3)-1$
$S_4 = \frac{\ln(8)}{2}-1$
$S_5 = \ln(\frac{5}{8} + \frac{\sqrt{5}}{8})(\sqrt{5}+1)-\sqrt{5}\ln(\frac{5}{8}-\frac{\sqrt{5}}{8}) + \frac{1}{10}[\ln(\frac{5}{8}-\frac{\sqrt{5}}{8}) + 4\ln(10)-10]$
$S_6 = \frac{\ln(432)}{6}-1$
My interest is piqued and i would love to see a generalized form of what $S_n$ would yield and how one would go about finding this.
 [/INST] Thanks to Raymond Manzoni's answer, i have found the generalized form
Since 
$$\psi(1+x) = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k(k+x)}$$
$$
\psi(1+\frac{1}{n})+\psi(1-\frac{1}{n})+2\gamma = \sum_{k=1}^{\infty}\frac{1}{nk(k+\frac{1}{n})}-\frac{1}{nk(k-\frac{1}{n})}
$$
$$
= \sum_{k=1}^{\infty}\frac{-2k}{n^2k^4-k^2} = \sum_{k=1}^{\infty}\frac{-2}{n^2 k^3-k} \therefore
$$
$$
S_n=\sum_{k=1}^{\infty}\frac{2}{n^3k^3-nk} = -\frac{\psi(1+\frac{1}{n})+\psi(1-\frac{1}{n})+2\gamma}{n}
$$
</s>