Show $\lim_{x \to0^+} \sum_{n=1}^{\infty} \frac{2x}{n^2x^2+1} = \pi$ Show that:
$$\lim_{x \to0^+} \sum_{n=1}^{\infty} \frac{2x}{n^2x^2+1} = \pi$$
 A: 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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$\ds{\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} = \pi:
     \ {\large ?}}$.

\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 $\ds{\lim_{x \to 0^{+}}\coth\pars{\pi \over x} = 1}$.

