# Find the sum of $\sum_{k=0}^{\infty} 1/[(r+2k+2)(s+2k+2)]$

\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.

• Do you know about partial fractions? – abiessu Oct 14 '18 at 4:47
• Not really, but I will look it up. – 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, $$\sum_{m=1}^\infty\frac1{(r+2m)(s+2m)}=\frac{\psi(r/2+1) -\psi(s/2+1)}{2(r-s)}.$$
• If $r$ and $s$ are integers of different parity, you can reduce to the case $(r,s)=(2,1)$, which is a standard series. – Lord Shark the Unknown Oct 14 '18 at 5:06
• @Pinteco. Use harmonic numbers instead if you prefer $$\sum_{m=1}^\infty\frac1{(r+2m)(s+2m)}=\frac{\psi(r/2+1) -\psi(s/2+1)}{2(r-s)}=\frac{H_{\frac{r}{2}}-H_{\frac{s}{2}}}{2( r- s)}$$ – Claude Leibovici Oct 14 '18 at 5:06