Evaluating series using real methods

**SworD** · August 29th 2013, 11:55 AM

It can be shown using complex analysis that,

$\sum_{n=1}^{\infty} \frac{1}{n^2+1} = \frac{\pi}{2}\left(\frac{e^{-\pi}+e^{\pi}}{e^{\pi}-e^{-\pi}}\right)-\frac{1}{2}$

Since this is a statement about real numbers, one would think that it is attainable using purely real methods, Ramanujan-style. Anyone with any ideas how to do so?

**FelixFelicis28** · September 3rd 2013, 02:01 AM

Originally Posted by SworD

It can be shown using complex analysis that,

$\sum_{n=1}^{\infty} \frac{1}{n^2+1} = \frac{\pi}{2}\left(\frac{e^{-\pi}+e^{\pi}}{e^{\pi}-e^{-\pi}}\right)-\frac{1}{2}$

Since this is a statement about real numbers, one would think that it is attainable using purely real methods, Ramanujan-style. Anyone with any ideas how to do so?

It doesn't use completely real techniques but no complex analysis.

$\displaystyle \frac{\sin x}{x} = \prod_{r \geq 1} \left(1 - \frac{x^2}{(r \pi)^2}\right) \implies \frac{\sin (i \pi x)}{i \pi x} = \prod_{r \geq 1} \left(1 + \frac{x^2}{r^2}\right)$

Taking logs of both sides:

$\ln \left( \frac{\sin (i \pi x)}{i \pi x} \right) = \sum_{r \geq 1} \ln \left( 1 + \frac{x^2}{r^2} \right) = \ln \left( \frac{\sinh (\pi x)}{\pi x} \right)$

Differentiating both sides wrt $x$

$\displaystyle \begin{aligned} \sum_{r \geq 1} \frac{2x}{r^2 + x^2} & = \frac{d}{dx} \ln \left( \frac{\sinh (\pi x)}{\pi x} \right) = \frac{1}{x} \Big( \pi x \coth (\pi x) - 1 \Big) \\ \implies \sum_{r \geq 1} \frac{1}{r^2 + x^2} & = \frac{1}{2x^2} \Big( \pi x \coth (\pi x) - 1 \Big) \end{aligned}$

Thus:

$\sum_{r \geq 1} \frac{1}{r^2 + 1} = \frac{1}{2x^2} \Big( \pi x \coth (\pi x) - 1 \Big) \bigg|_{x = 1} = \frac{1}{2} \Big( \pi \coth \pi - 1 \Big) = \frac{1}{2} \left( \pi \cdot \frac{e^{\pi} + e^{-\pi}}{e^{\pi} - e^{-\pi}} - 1 \right)$

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