How to calculate $\lim_{x\to0} \sum_{n=1}^\infty \frac{\sin x}{4+n^4x^4}$ using calculus I have problem with calculating this limit, I suppose i should transform it into some integral but I don't know how:
$$
\lim_{x\to0} \sum_{n=1}^\infty \frac{\sin x}{4+n^4x^4}
$$
 A: As written the limit does not exist. However, by considering the logarithmic derivative of the Weierstrass product for the sine function we have
$$ \cot z = \frac{1}{z}+\sum_{n\geq 1}\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right)\tag{1} $$
hence by partial fraction decomposition it follows that
$$ \sum_{n\geq 1}\frac{z}{4+n^2 z^2} = \frac{1+i}{16} \left[(i-1) z+\pi  \cot\left(\frac{(1+i) \pi }{z}\right)+\pi\,\text{coth}\left(\frac{(1+i) \pi }{z}\right)\right] \tag{2} $$
and
$$ \lim_{z\to 0^{\pm}} \sum_{n\geq 1}\frac{\sin z}{4+n^2 z^2} = \pm\frac{\pi}{8}\tag{3} $$
as previously stated.
A: I suppose that $x>0$ and $x\to 0$. As $\sin(x)/x \to 1$, it is sufficient to study
$\displaystyle S(x)=\sum_{k\geq 1}\frac{x}{4+(kx)^4}$.
Now, for $k\leq t\leq k+1$, we have
$$\frac{x}{4+(k+1)^4x^4}\leq \frac{x}{4+t^4x^4}\leq \frac{x}{4+k^4x^4} $$
And integrating from $k$ to $k+1$ and adding we get that 
$$S(x)-\frac{x}{4+x^4}\leq \int_1^{+\infty}\frac{x}{4+t^4x^4}dt\leq S(x)$$
Now it is sufficient to study $\int_1^{+\infty}\frac{x}{4+t^4x^4}dt$, or even as it is clear that $\int_0^{1}\frac{x}{4+t^4x^4}dt\to 0$ as $x\to 0$, $\int_0^{+\infty}\frac{x}{4+t^4x^4}dt$, and this last integral is $\int_0^{+\infty}\frac{du}{4+u^4}$ by the change of variable $u=tx$. Hence the limit as $x\to 0$, $x>0$ is $L=\int_0^{+\infty}\frac{du}{4+u^4}$. It remains to compute the integral....
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{x\ \to\ 0^{\color{#f00}{\large\,\pm}}}
\sum_{n = 1}^{\infty}{\sin\pars{x} \over 4 + n^{4}x^{4}}:\ {\large ?}}$.

\begin{align}
\mbox{Note that}\quad
\lim_{x\ \to\ 0^{\color{#f00}{\large\,\pm}}}
\sum_{n = 1}^{\infty}{\sin\pars{x} \over 4 + n^{4}x^{4}} & =
{1 \over 4}\,\lim_{x\ \to\ \color{#f00}{\pm\infty}}\bracks{%
x^{4}\sin\pars{\root{2} \over x}\sum_{n = 1}^{\infty}{1 \over n^{4} + x^{4}}}
\\[5mm] & =
{1 \over 4}\,\lim_{x\ \to\ \color{#f00}{\pm\infty}}\bracks{%
x^{2}\sin\pars{\root{2} \over x}
\bbox[15px,#ffe]{\ds{\Im\sum_{n = 1}^{\infty}{1 \over n^{2} -\ic x^{2}}}}}
\label{1}\tag{1}
\end{align}

\begin{align}
\bbox[15px,#ffe]{\ds{\Im\sum_{n = 1}^{\infty}{1 \over n^{2} -\ic x^{2}}}} & =
\Im\sum_{n = 0}^{\infty}{1 \over
\pars{n + 1 - \expo{\ic\pi/4}\verts{x}}
\pars{n + 1 + \expo{\ic\pi/4}\verts{x}}}
\\[5mm] & =
\Im\bracks{\Psi\pars{1 + \expo{\ic\pi/4}\verts{x}} -
\Psi\pars{1 - \expo{\ic\pi/4}\verts{x}} \over 2\expo{\ic\pi/4}\verts{x}}\quad
\pars{\substack{\Psi:\ Digamma\ Function}}
\end{align}
By replacing in \eqref{1}:
\begin{align}
\lim_{x\ \to\ 0^{\color{#f00}{\large\,\pm}}}
\sum_{n = 1}^{\infty}{\sin\pars{x} \over 4 + n^{4}x^{4}} & =
\pm\,{\root{2} \over 8}\lim_{x\ \to\ \color{#f00}{\infty}}
\Im\braces{\expo{-\ic\pi/4}\bracks{\Psi\pars{1 + \expo{\ic\pi/4}\verts{x}} -
\Psi\pars{1 - \expo{\ic\pi/4}\verts{x}}}}
\\[5mm] & \substack{\Psi\ Recursion\\[1mm] {\large =}}\,\,\,
\pm\,{\root{2} \over 8}\lim_{x\ \to\ \color{#f00}{\infty}}
\Im\braces{\expo{-\ic\pi/4}\bracks{\Psi\pars{\expo{\ic\pi/4}\verts{x}} -
\Psi\pars{\expo{-3\ic\pi/4}\verts{x}}}}
\end{align}


Note that
  $\ds{\Psi\pars{z} \sim
\ln\pars{z} - {1 \over 2z} + \,\mrm{O}\pars{1 \over z^{2}}\
\mbox{as}\ \verts{z} \to \infty\ \mbox{with}\ \verts{\mrm{arg}\pars{z}} < \pi}$.

Then,
$$
\lim_{x\ \to\ 0^{\color{#f00}{\large\,\pm}}}
\sum_{n = 1}^{\infty}{\sin\pars{x} \over 4 + n^{4}x^{4}} =
\pm\,{\root{2} \over 8}\,\Im\pars{\expo{-\ic\pi/4}\pi\ic} =
\bbx{\pm\,{\pi \over 8}}
$$
A: The limit is not well-defined.
If $x>0$ and $x \to 0$, then the limit is $\frac{1}{8}\pi$.
If $x<0$ and $x \to 0$, then the limit is $-\frac{1}{8}\pi$.
A: For $x>0$ , integration from 0 to 1,      $(x/5)sin (1/x)$
For $x <0$ integration from 0 to 1  ,         $-(x/5)sin (1/x)$
