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Show that:

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

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2 Answers 2

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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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  • $\begingroup$ Now I see, thank you : ) $\endgroup$
    – Prold
    Mar 30, 2014 at 15:39
  • $\begingroup$ Great, glad to help $\endgroup$ Mar 30, 2014 at 22:35
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} = \pi: \ {\large ?}}$.

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

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