\begin{align} \sum_{k=0}^{\infty} \frac{1}{(r+2k+2)(s+2k+2)} = \sum_{m=1}^{\infty} \frac{1}{(r+2m)(s+2m)} \end{align}

$r>s>0$. $r$ is a odd number, but I don't think this gonna be relevant. The fact that $r>s$ is given because it's a telescopic series (I think).

I wanted to know where does this series converge to and what happens when $r$ is odd and $s$ is either odd or even, but I don't know how to evaluate this sum.

  • 1
    $\begingroup$ Do you know about partial fractions? $\endgroup$ – abiessu Oct 14 '18 at 4:47
  • $\begingroup$ Not really, but I will look it up. $\endgroup$ – Pinteco Oct 14 '18 at 4:55

$$\frac1{(r+2m)(s+2m)}=\frac1{r-s}\left(\frac1{s+2m}-\frac1{r+2m}\right).$$ If the difference of $r$ and $s$ is an even integer, the sum will telescope. In general, if $r-s$ is odd, or not an integer, this won't work.

In these cases use the identity involving the digamma function $$\frac{1}{x}=\psi(x+1)-\psi(x).$$ One gets $$\sum_{m=1}^M\frac1{(r+2m)(s+2m)}=\frac{\psi(s/2+M+1)-\psi(s/2+1) -\psi(r/2+M+1)+\psi(r/2+1)}{2(r-s)}.$$ In the limit, $$\sum_{m=1}^\infty\frac1{(r+2m)(s+2m)}=\frac{\psi(r/2+1) -\psi(s/2+1)}{2(r-s)}.$$

  • $\begingroup$ Is there another way rather than using the digamma function? $\endgroup$ – Pinteco Oct 14 '18 at 5:02
  • $\begingroup$ If $r$ and $s$ are integers of different parity, you can reduce to the case $(r,s)=(2,1)$, which is a standard series. $\endgroup$ – Lord Shark the Unknown Oct 14 '18 at 5:06
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    $\begingroup$ @Pinteco. Use harmonic numbers instead if you prefer $$\sum_{m=1}^\infty\frac1{(r+2m)(s+2m)}=\frac{\psi(r/2+1) -\psi(s/2+1)}{2(r-s)}=\frac{H_{\frac{r}{2}}-H_{\frac{s}{2}}}{2( r- s)}$$ $\endgroup$ – Claude Leibovici Oct 14 '18 at 5:06

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