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How to prove that this series converges? $\exists\space\text{s}\space\wedge\space\exists\space\omega\in\mathbb{C}$:

$$\sum_{\text{n}=0}^\infty\frac{1}{\text{s}^2+\left(1+2\text{n}\right)^2\omega^2}$$

asdasd
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3 Answers3

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Note that we can also calculate the closed form of this series. We have $$\sum_{n\in\mathbb{Z}}\frac{1}{z^{2}+(1+2n)^{2}}=2\sum_{n\geq0}\frac{1}{z^{2}+\left(1+2n\right)^{2}}$$ and now we can use the well known summation formula $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\cot\left(\pi w\right)f\left(w\right)\textrm{ at }f\left(w\right)\textrm{'s poles}\right\} $$ and now since we have poles at $\frac{1}{2}i\left(z\pm i\right) $ we get $$-\textrm{Res}_{w=-\frac{i\left(z-i\right)}{2}}\pi\cot\left(\pi w\right)f\left(w\right)-\textrm{Res}_{w=\frac{i\left(z+i\right)}{2}}\pi\cot\left(\pi w\right)f\left(w\right)=\frac{\pi\tanh\left(\frac{\pi z}{2}\right)}{2z} $$ hence $$\sum_{n\geq0}\frac{1}{z^{2}+\left(1+2n\right)^{2}}=\frac{\pi\tanh\left(\frac{\pi z}{2}\right)}{4z}$$ now taking $z=s/\omega $ we have $$\sum_{n\geq0}\frac{1}{s^{2}+\left(1+2n\right)^{2}\omega^{2}}=\color{green}{\frac{\pi\tanh\left(\frac{\pi s}{2\omega}\right)}{4\omega s}}.$$for $\omega\neq0$ and $s\neq0$.

Marco Cantarini
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  • When I asked Mathematice I got that the sum equals $\frac{\pi\tanh\left(\frac{\pi\text{s}}{2\omega}\right)}{4\text{s}\omega}$. So you got the same as me (for the tanh part) but you've $-\frac{1}{\text{s}^2+\omega^2}$ extra. See http://math.stackexchange.com/questions/2022848/infinite-sum-and-the-conditions-with-tanh – asdasd Nov 21 '16 at 13:38
  • @asdasd My series starts from $n=1.$ – Marco Cantarini Nov 21 '16 at 13:44
  • I dont actually see what you mean, so can you maybe write the same proof but then with the starting point $\text{n}=0$? – asdasd Nov 21 '16 at 14:08
  • @asdasd The "extra term" is only the value of the series at $n=0$. For some unknown reason I read that $n$ starts from $1$ instead of $0$ in your question. The proof is the same, you have only to observe that $$\sum_{n\in\mathbb{Z}}\frac{1}{z^{2}+(1+2n)^{2}}=2\sum_{n\geq0}\frac{1}{z^{2}+\left(1+2n\right)^{2}}.$$ – Marco Cantarini Nov 21 '16 at 14:11
  • I see it, but when you've the time and if you want to help me can you rewrite your answer,by starting at $\text{n}=0$?! Thanks for your help anyway – asdasd Nov 21 '16 at 14:23
  • @asdasd I changed my answer, but I think you have not understood my proof. – Marco Cantarini Nov 21 '16 at 14:27
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As $\;s.w\in\Bbb C\;$ , we can write:

$$\left|\frac1{s^2+(1+2n)^2w^2}\right|=\frac1{|s^2+(1+2n)^2w^2|}\le\frac1{(1+2n)^2|w|^2-|s|^2}$$

and since the series with the rightmost general term converges then so does absolutely your series...of course, assuming $\;w\neq0\;$ .

DonAntonio
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I suppose that $ \omega \ne 0$.

We have

$\frac{1}{\text{s}^2+\left(1+2\text{n}\right)^2\omega^2} \le \frac{1}{4n^2 \omega^2}$.

Fred
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