0
$\begingroup$

\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.

$\endgroup$
  • 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
1
$\begingroup$

$$\frac1{(r+2m)(s+2m)}=\frac1{r-s}\left(\frac1{s+2m}-\frac1{r+2m}\right).$$ 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, $$\sum_{m=1}^\infty\frac1{(r+2m)(s+2m)}=\frac{\psi(r/2+1) -\psi(s/2+1)}{2(r-s)}.$$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.