Asymptotic behavior of $\sum\limits_{n=1}^{\infty} \frac{nx}{(n^2+x)^2}$ when $x\to\infty$ Prove that : $$\lim_{x\to \infty} \sum_{n=1}^{\infty} \frac{n x}{(n^2+x)^2} = \frac{1}{2}.$$ Furthermore, can we have an asymptotic expansion here? Mathematica gives a pretty one, but I could not prove it.

Doing an Asymptotic expansion for the summand gives that it like $\frac{n}{x}$ for large $x$ but this is completely useless, what I'm I messing here? 
I prefer that you do not write the full solution if you feel that it is easy.
 A: For every $n\geqslant1$, let $u_n(x)=\dfrac{nx}{(n^2+x)^2}$ and $v_n(x)=\dfrac{x}{n^2+x}$, then 
$$
v_n(x)-v_{n+1}(x)=\dfrac{x(2n+1)}{(n^2+x)((n+1)^2+x)},
$$ 
hence
$$
\frac{2n+1}{n+1}u_{n+1}(x)\leqslant v_n(x)-v_{n+1}(x)\leqslant\frac{2n+1}nu_n(x).
$$
Thus, for every $n\geqslant1$,
$$
\frac{n}{2n+1}v_n(x)\leqslant\sum_{k=n}^{+\infty}u_k(x)\leqslant\frac{n}{2n-1}v_{n-1}(x).
$$
Since $v_n(x)\to1$ and $v_{n-1}(x)\to1$ when $x\to+\infty$,
$$
\frac{n}{2n+1}\leqslant\liminf_{x\to\infty}\sum_{k=n}^{+\infty}u_k(x)\leqslant\limsup_{x\to\infty}\sum_{k=n}^{+\infty}u_k(x)\leqslant\frac{n}{2n-1}.
$$
For each $k\geqslant1$, $u_k(x)\to0$ when $x\to\infty$, hence
$$
\frac{n}{2n+1}\leqslant\liminf_{x\to\infty}\sum_{k=1}^{+\infty}u_k(x)\leqslant\limsup_{x\to\infty}\sum_{k=1}^{+\infty}u_k(x)\leqslant\frac{n}{2n-1}.
$$
This holds for every $n\geqslant1$ and the leftmost and rightmost factors both converge to $\frac12$ when $n\to\infty$, hence
$$
\lim_{x\to\infty}\sum_{k=1}^{+\infty}u_k(x)=\frac12.
$$
A: Substituting $x=\frac1{u^2}$ and using a Riemann Sum, we can compute
$$
\begin{align}
\lim_{x\to\infty}\sum_{n=1}^\infty\frac{nx}{(n^2+x)^2}
&=\lim_{u\to0}\sum_{n=1}^\infty\frac{un}{(u^2n^2+1)^2}u\\
&=\int_0^\infty\frac{t}{(t^2+1)^2}\,\mathrm{d}t\\[3pt]
&=\frac12
\end{align}
$$
Since $\frac{t}{(t^2+1)^2}$ increases from $0$ to $\frac{3\sqrt3}{16}$ on $\small\left[0,\frac1{\sqrt3}\right]$ and decreases from $\frac{3\sqrt3}{16}$ to $0$ on $\small\left[\frac1{\sqrt3},\infty\right)$,
the difference between the Riemann Sum and the integral should be at most
$$
\frac{3\sqrt3}{8}\,u=\frac{3\sqrt3}{8}\,x^{-1/2}
$$
