converge value of series $\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) $ \begin{align}
\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) = \sum_{n=0}^{\infty}\frac{4d}{(n+d+1)(n+5d+1)}=
?
\end{align}
I know from the $p$-test, ($i.e$ $\sum \frac{1}{n^p}$ : $p>1$ series converges)
The above series converges. 
I want to know the exact value(or function) in terms of $d$. 

One may recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$,
$$
\psi(u+1) = -\gamma + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{u+n}  
\right), \quad u >-1, \tag1
$$ where $\gamma$ is the Euler-Mascheroni. From $(1)$ you get
$$
\sum_{n= 1}^{N}\frac{1}{n+u}=\psi(N+u+1) -\psi(u+1) .
$$
Assume $d$ is any real number such that $d>-1/5$. You may write, for $N\geq1$,
$$
\begin{align}
\sum_{n= 1}^{N}\left(\frac{1}{n+d+1} - \frac{1}{n+5d+1}\right)&=\sum_{n= 1}^{N}\frac{1}{n+d+1} -\sum_{n= 1}^{N} \frac{1}{n+5d+1}\\\\
&=\left(\psi(N+d+2)-\psi(d+2)\right)-(\psi(N+5d+2)-\psi(5d+2))
\end{align}
$$ Then letting  $N \to \infty$, using $\displaystyle \psi(M)=\log M-O(1/M)$ as $M \to +\infty$, gives

$$
\begin{align}
\sum_{n= 1}^{\infty}\left(\frac{1}{n+d+1} - \frac{1}{n+5d+1}\right)&=\psi(5d+2)-\psi(d+2).
\end{align}
$$ 

Many special values of $\psi$ are known, for example
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac13\right) & =  -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3.
\end{align}
$$
