<s>[INST] Asymptotic expansion in 0 of a series I need to find a 2 term asymptotic expansion of the serie of functions $f=\sum u_n$
where for all real $x$, $u_n(x) = 1/(n^2+x^2)$. Obviously, the function being continuous (which is easy to prove by uniform convergence over $\mathbb R$) the limit is $\pi^2/6$.
I have no idea on what to do next. I have tried to get an inequality by using serie integral comparison: if we take a real $x$, then define the function $f : t\in \mathbb R \rightarrow 1/(t^2+x^2)$, which is positive and decreasing over $\mathbb R_+$, we can get:
(skipping the integration part)
$$\arctan(x)/x -1 +1/(1+x^2)-1 \le f(x) - \pi^2/6 \le \arctan(x)/x -1$$
which gives no useful information. Any help would be appreciated.
 [/INST] According to Find the sum of $\sum \frac{1}{k^2 - a^2}$ when $0<a<1$, the following identity holds
$$\sum_{n=1}^\infty \frac1{n^2-z^2}=\frac1{2z^2}-\frac{\pi\cot\,\pi z}{2z}.$$
After letting $z=ix$, we find
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\frac{\pi\coth(\pi x)}{2x}-\frac1{2x^2}.$$
Finally, after expanding the Hyperbolic Cotangent at $x=0$, we get
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\frac{\pi^2}{6}-\frac{\pi^4}{90}x^2 +O(x^4).$$
Notice that $\frac{\pi^4}{90}=\sum_{n=1}^{\infty}\frac{1}{n^4}$:
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\sum_{n=1}^\infty \frac{1}{n^2}\frac1{1+(x/n)^2}=\sum_{n=1}^\infty \frac{1}{n^2}\left(1-\frac{x^2}{n^2}+O(x^4)\right)\\=\sum_{n=1}^\infty\frac1{n^2}-\sum_{n=1}^\infty\frac1{n^4}\,x^2+O(x^4).$$
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