Find $\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}$ I have a problem with finding the following limit. I suspect that it should be easy, but really I don't have a clue.
$$
\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}
$$
 A: Note that if $(n-1)x \leqslant u \leqslant nx \leqslant v \leqslant (n+1)x,$ then
$$\frac{1}{4 + v^2} \leqslant  \frac{1}{4 +n^2x^2} \leqslant \frac{1}{4 + u^2}, $$
and
$$\int_{nx}^{(n+1)x}\frac{dv}{4 + v^2} \leqslant  \frac{x}{4 +n^2x^2} \leqslant \int_{(n-1)x}^{nx}\frac{du}{4 + u^2}, $$
Summing we get
$$\int_x^\infty \frac{dv}{4 + v^2} \leqslant \sum_{n=1}^\infty \frac{x}{4 +n^2x^2} \leqslant \int_0^\infty \frac{du}{4 + u^2} .$$
Using the squeeze principle,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{x}{4 +n^2x^2} = \int_0^\infty \frac{du}{4 + u^2}.$$
Hence,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{\sin x}{4 +n^2x^2} =  \lim_{x \to 0} \frac{\sin x}{x} \sum_{n=1}^\infty \frac {x}{4 +n^2x^2} \\ = \int_0^\infty \frac{du}{4 + u^2}$$
A: Without justifying any single detail, I propose the following solution. I introduce the evocative symbol
$$
\Delta t:=x
$$
and define the sequence $t_n=n\Delta t$. Then
$$
\lim_{x\to0}\sum_{n=1}^{\infty}\frac{sinx}{4+n^2x^2}=\lim_{x\to0}\sum_{n=1}^{\infty}\frac{x}{4+n^2x^2}=\lim_{\Delta t\to0}\sum_{t_n\in \Delta t\cdot \mathbb N}\frac{\Delta t}{4+t_n^2}=\int_0^\infty\frac{dt}{4+t^2}=\frac{\pi}{4}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 1}^{N}{1 \over 4 + n^{2}x^{2}} & =
{1 \over x^{2}}\sum_{n = 1}^{N}{1 \over n^{2} + 4/x^{2}} =
{1 \over 2x}\,\Im\sum_{n = 0}^{N - 1}{1 \over n + 1 - 2\ic/x}
\\[5mm] & =
{1 \over 2x}\,\Im\sum_{n = 0}^{\infty}\pars{{1 \over n + 1 - 2\ic/x} -
{1 \over n + N + 1 - 2\ic/x}}
\\[5mm] & =
{1 \over 2x}\,\Im\pars{H_{N - 2\ic/x} - H_{-2\ic/x}}
\qquad\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & \stackrel{N\ \to\ \infty}{\to}\,\,\,
-\,{1 \over 2}\,\Im\pars{H_{-2\ic/x} \over x}
\end{align}


With $\ds{\lim_{x \to 0}{\sin\pars{x} \over x} = 1}$:

\begin{align}
\lim_{x \to 0}\sum_{n = 1}^{\infty}{\sin\pars{x} \over 4 + n^{2}x^{2}} & =
-\,{1 \over 2}\,\lim_{x \to 0}\Im\pars{H_{-2\ic/x}} =
-\,{1 \over 2}\,\lim_{x \to 0}\Im\,\Psi\pars{1 - {2 \over x}\,\ic} =
-\,{1 \over 2}\,\lim_{x \to 0}\arctan\pars{-\,{2 \over x}}
\end{align}

where $\ds{\Psi}$ is the Digamma Function. Note that
  $\bbx{\ds{\lim_{x \to 0^{\color{red}{\Large\pm}}}\sum_{n = 1}^{\infty}
{\sin\pars{x} \over 4 + n^{2}x^{2}} = \color{red}{\pm}{\pi \over 4}}}$.

