1

I'm working on a problem, and I'm stuck in the calculations of finding

$\sum\limits_{n=0}^{\infty}\frac{1}{1+n^2}$

Suggestions on how to approach this calculation? Thanks!

(Also, I used Fourier to get to this result from where we were... so using the Fourier somehow feels counter intuitive, but I might be wrong)

Thanks in advance.

Martin Sleziak
  • 53,687
phoenix
  • 671
  • Hint : Let $f(z) := \frac{1}{z^2+1}$. Note that $|z f(z)| \to 0$ as $|z| \to \infty$. Show that $g(z) := f(z) \cot \pi z$ has simple poles at $\pm i$ and that $\text{res}(g, \pm i) = - \frac{\pi \coth \pi}{2}$. Then conclude that the sum equals $\frac{\pi \coth \pi+1}{2}$ – Amateur Apr 07 '14 at 00:15
  • 3
    See also ZZ's upvoted comment here. – Lucian Apr 07 '14 at 00:33
  • This question http://math.stackexchange.com/questions/1666859/the-sum-of-the-series-sum-limits-n-0-infty-frac1n23 is very similar. – Jack D'Aurizio Feb 22 '16 at 11:26

1 Answers1

3

The identity $$ \sum_{n\geq 0}\frac{1}{n^2+\alpha^2} = \frac{1}{2\alpha^2}+\frac{\pi}{2\alpha}\cdot\coth(\pi\alpha) \tag{1}$$ that holds for any $\alpha>0$, can be proved through the Weierstrass product for the sine function and the properties of the logarithmic derivative, just like here. For instance, by considering that $$ \lim_{\alpha\to 0^+}\left(-\frac{1}{2\alpha^2}+\frac{\pi}{2\alpha}\cdot\coth(\pi\alpha)\right) = \frac{\pi^2}{6}\tag{2}$$ we may also easily compute $\zeta(2)$. Another approach for proving $(1)$ relies on: $$ \int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-x}\,dx = \frac{1}{n^2+1}\tag{3}$$ and the fact that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of a sawtooth wave, giving: $$ \sum_{n\geq 1}\frac{1}{n^2+1} = \frac{e^{2\pi}}{e^{2\pi}-1}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx\tag{4} $$ from which:

$$ \color{red}{\sum_{n\geq 0}\frac{1}{n^2+1}} = \frac{1+\pi\coth(\pi)}{2} = \color{red}{\frac{1}{2}+\frac{\pi}{2}\cdot\frac{e^{2\pi}+1}{e^{2\pi}-1}} \tag{5}$$

easily follows.

Community
  • 1
Jack D'Aurizio
  • 353,855