Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that:

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

share|cite|improve this question
you might find this link helpful:… ,divide left side by $z^2$ – Ziqian Xie Mar 30 '14 at 14:54
up vote 7 down vote accepted

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.)

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

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} = \pi: \ {\large ?}}$.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.