Find the value of $$\sum^{n-1}_{m=1}\left(\frac{1}{n-m}+\frac{1}{n+m}\right)$$ I used WolframAlpha obtaining $$\psi^{(0)}(2n)-\frac{1}{n}+\gamma$$ where $\gamma$ is the Euler-Mascheroni constant and $\psi^{(k)}(x)$ is the k-th derivative of the digamma function. But I can not prove it. Any suggestions please?

  • $\begingroup$ The $\psi^{(0)}$ formula is a mere rewriting, mainly useless. You might want to use $$\sum^{n-1}_{m=1}\left(\frac{1}{n-m}+\frac{1}{n+m}\right)=\sum^{n-1}_{k=1}\frac{1}{k}+\sum^{2n-1}_{i=n+1}\frac{1}{i}=\sum^{2n-1}_{m=1}\frac{1}{m}-\frac{1}{n}-\frac{1}{n+1}.$$ $\endgroup$ – Did May 24 '15 at 9:04
  • $\begingroup$ please fix you $\LaTeX$ $\endgroup$ – Dr. Sonnhard Graubner May 24 '15 at 9:27

Your sum is just $H_{2n-1}-\frac{1}{n}$, where $H_n$ is the $n$-th harmonic number, defined as: $$ H_n = \sum_{k=1}^{n}\frac{1}{k}.$$ $H_n$ can be defined also for non-integer $n$s, since $f(x)=\frac{d}{dx}\log\Gamma(x+1)=\psi(x+1)$ satisfies the recursive relation $f(x+1)=f(x)+\frac{1}{x}$. So $\psi(2n)+\gamma$ is just a re-writing for $H_{2n-1}$.


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