\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 $d$.

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    $\begingroup$ $$\sum_{k=d+1}^{5d}\frac1k=H_{5d}-H_d$$ $\endgroup$ – Did May 15 '15 at 10:25

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 $$ \begin{align} \psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\ \psi \left(\frac13\right) & = -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3. \end{align} $$

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  • $\begingroup$ Actually the above series comes from digamma function that you answered. What i want to do is compute, see : here. $\endgroup$ – phy_math May 15 '15 at 12:49

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