# Show $\lim_{x \to0^+} \sum_{n=1}^{\infty} \frac{2x}{n^2x^2+1} = \pi$

Show that:

$$\lim_{x \to0^+} \sum_{n=1}^{\infty} \frac{2x}{n^2x^2+1} = \pi$$

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you might find this link helpful: math.stackexchange.com/questions/3525/… ,divide left side by $z^2$ –  Ziqian Xie Mar 30 '14 at 14:54

Rename $x$ as $\Delta x$. That may help you to see that your limit is $\int_0^\infty{f(x)\,dx}$, where $f(x) = {2\over x^2+1}$. (You should recognize your limit as the limit of a Riemann sum.)

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Now I see, thank you : ) –  Prold Mar 30 '14 at 15:39
Great, glad to help –  Jason Zimba Mar 30 '14 at 22:35


\begin{align} \lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} &=2\lim_{x \to 0^{+}}\bracks{{1 \over x} \sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + \ic/x}\pars{n + 1 - \ic/x}}} \\[3mm]&=2\lim_{x \to 0^{+}}\bracks{{1 \over x} \,{\Psi\pars{1 + \ic/x} - \Psi\pars{1 - \ic/x}\over \pars{1 + \ic/x} - \pars{1 - \ic/x}}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf 6.3.1}$ and we used the identity ${\bf 6.3.16}$.

\begin{align} \lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} &=2\,\lim_{x \to 0^{+}}\Im\Psi\pars{1 + {\ic \over x}} \end{align}

With the identity ${\bf 6.3.13}$: \begin{align} \color{#66f}{\large\lim_{x \to 0^{+}} \sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1}} &=\lim_{x \to 0^{+}}\bracks{-x + \pi\coth\pars{\pi \over x}} =\color{#66f}{\LARGE\pi} \end{align} since $\ds{\lim_{x \to 0^{+}}\coth\pars{\pi \over x} = 1}$.

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