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Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ , where $ \ l=constant \ $

Answer

The given series is convergent clearly .

$ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \\ = \frac{1}{1^2+l^2}+\frac{1}{2^2+l^2}+......+\frac{1}{n^2+l^2}+........ \\ = \frac{1}{l} [ d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{n}{l})+.....] \\ = \frac{1}{l} \lim_{n \to \infty} \sum_{k=1}^{n}[d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{k}{l})] \\ = \frac{1}{l} \ d (\lim_{n \to \infty} \sum_{k=1}^{n}[ \tan^{-1}(\frac{1}{l})+ \tan^{-1}(\frac{2}{l})+........+ \tan^{-1}(\frac{k}{l})]) $

but now i can't proceed to find the sum of the series.

Help me out.

MAS
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3 Answers3

5

Let $f(z)=\frac{\cot(\pi z)}{z^2+\ell^2}$. Note that $f$ has simple poles at $z=n$ and $z=\pm i\ell$.

Next, let $R_N=(N+1/2)$ in $(1)$. Then, the residue theorem guarantees that

$$\oint_{|z|=R_N}f(z)\,dz=2\pi i \sum_{n=-N}^N\text{Res}\left(\frac{\cot(\pi z)}{z^2+\ell^2}, z=n\right)+2\pi i \text{Res}\left(\frac{\cot(\pi z)}{z^2+\ell^2}, z=\pm i\ell\right)\tag 1$$

In THIS ANSWER, I showed that $|\cot(\pi z)|$ is bounded on the circle $|z|=N+1/2$, where $N\in \mathbb{N}$.

So, letting $N\to \infty$, the contour integral in $(1)$ tends to $0$ and we find

$$\begin{align} \sum_{n=-\infty }^\infty \frac{1}{\pi(n^2+\ell^2)}&=-\left(\frac{\cot(i\pi\ell)}{i2\ell}+\frac{\cot(-i\pi\ell)}{-i2\ell}\right)\\\\ &=\frac{\coth(\pi\ell)}{\ell}\tag2 \end{align}$$

Therefore, multiplying $(2)$ by $\pi$ reveals

$$\sum_{n=-\infty }^\infty \frac{1}{n^2+\ell^2}=\frac{\pi \coth(\pi\ell)}{\ell}$$

Finally, exploiting the evenness of the summands yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1 }^\infty \frac{1}{\pi(n^2+\ell^2)}=\frac{\pi\coth(\pi\ell)}{2\ell}-\frac{1}{2\ell^2}}$$

Mark Viola
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3

Hint: Use the fact that \begin{align} \sum^\infty_{n=-\infty} \frac{1}{x^2+n^2} = \frac{\pi\coth(\pi x) }{x}. \end{align}

Jacky Chong
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  • 1
    how to prove that your result ? – MAS Nov 21 '17 at 03:27
  • @yourmath It can be derived from the infinite product for the sine function. – Angina Seng Nov 21 '17 at 03:29
  • @yourmath I guess that the LHS is the Fourier series of the RHS – Masacroso Nov 21 '17 at 03:30
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    @Masacroso No. Try showing $\frac{\pi\coth(\pi z) }{z}-\sum^\infty_{n=-\infty} \frac{1}{z^2+n^2}$ is a bounded entire function – reuns Nov 21 '17 at 03:36
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    While not quite the Fourier series, it can also be viewed as an expansion about the poles, a sort of infinite version of partial fractions: write it as a sum $\sum_k \frac{a_k}{x-b_k}$, where $b_k$ are the poles of the RHS and $a_k$ are the residues; one finds that the $b_k$ are $\pm in$ (there are no real poles except the one at 0, which is a double pole) and then combines the terms to produce a real result. Of course, this is far from rigorous, but it provides a sort of justification... – Steven Stadnicki Nov 21 '17 at 03:37
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At equation $(7)$ of this answer, it is concluded that, for all $z\in\mathbb{C}\setminus\mathbb{Z}$, $$ \sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z) $$ Therefore, $$ \begin{align} \sum_{k=1}^\infty\frac1{k^2+z^2} &=-\frac1{2iz}\sum_{k=1}^\infty\left(\frac1{iz-k}+\frac1{iz+k}\right)\\ &=-\frac1{2z^2}-\frac1{2iz}\sum_{k=-\infty}^\infty\frac1{iz+k}\\ &=-\frac1{2z^2}-\frac1{2iz}\pi\cot(\pi iz)\\ &=\bbox[5px,border:2px solid #C0A000]{-\frac1{2z^2}+\frac1{2z}\pi\coth(\pi z)} \end{align} $$

robjohn
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