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Show that $$\sum_{n=-\infty}^\infty \frac{1}{(3n-1)^2} = \frac{4\pi^2}{27}$$

Here is what I tried so far. I know that I can use the Residue theorem to solve a summation of this form.

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$

So the poles of $\frac{1}{(3z-1)^2}$ is z= $\frac{1}{3}$ (order 2).

Next I need to find the residues of $\frac{1}{(3n-1)^2} \times \pi \cot(\pi z)$ at z= $\frac{1}{3}$. But, the limit of $\frac{1}{(3z-1)^2} \times \pi \cot(\pi z)$ at z= $\frac{1}{3}$ is $\infty$.

I am not sure what to do at this point. Any help is appreciated.

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The mistake is that you calculate the residue in the wrong way. Define $$g(z)=\frac{\pi\cot(\pi z)}{(3z-1)^2}=\frac{h(z)}{(z-\frac{1}{3})^2},$$ where $h(z)=\frac{\pi}{9}\cot(\pi z)$. Note that $z=\frac{1}{3}$ is a pole of order $2$ and $h$ is analytic at $\frac{1}{3}$ with $h(\frac{1}{3})\neq 0$. Thus $$\text{res}_{z=\frac{1}{3}}g(z)=h'(\frac{1}{3})=-\frac{\pi^2}{9}\csc^2(\pi z)|_{z=\frac{1}{3}}=-\frac{4\pi^2}{27}.$$

In general, if $z_0$ is a pole of order $m$ of the function $g$, then we can write $$g(z)=\frac{h(z)}{(z-z_0)^m}$$ near the point $z_0$ such that $h$ is analytic at $z_0$ with $h(z_0)\neq 0$. Moreover in this case we have $$\text{res}_{z=z_0}g(z)=\frac{h^{(m-1)}(z_0)}{(m-1)!}.$$

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  • $\begingroup$ Great answer. Thank you very much! $\endgroup$ – user340423 May 18 '16 at 5:54
  • $\begingroup$ Can we directly use residue theorem to sum this? Isn't that required that we have no singularities on Real axis? Here we definitly have one at $\frac{1}{3}$ $\endgroup$ – vanHohenheim Aug 27 '17 at 19:35
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Just for the fun of it since you did ask for this problem solution using complex analysis and here we don't:

$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(3n-2)^2}+\sum_{n=1}^\infty\frac1{(3n-1)^2}+\overbrace{\frac19\sum_{n=1}^\infty\frac1{n^2}}^{=\frac19\frac{\pi^2}6}\implies$$

$$\sum_{n=1}^\infty\frac1{(3n-2)^2}+\sum_{n=1}^\infty\frac1{(3n-1)^2}=\frac4{27}\pi^2$$

Now just observe that

$$\sum_{n=-\infty}^0\frac1{(3n-1)^2}=\sum_{n=1}^\infty\frac1{(3n-2)^2}$$

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