<s>[INST] Showing that $\sum_{n \geq 1} \frac{a}{n^2 + a^2} $ is bounded by integrals. For $ a > 0$, we have: $ S(a) = \sum_{n \geq 0} \frac{a}{n^2 + a^2} $
Prove that $ S(a) $ is bounded between 2 integrals.

If we take: $ k \leq x \leq k + 1 $, we will have:
$ S(k+1) \leq \frac{a}{n^2 + a^2} \leq S(k) $
$ \implies \int_{k}^{k+1} S(k+1) da \leq \int_{k}^{k+1}\frac{a}{n^2 + a^2} da \leq \int_{k}^{k+1} S(k) da $
$ \implies \int_{0}^{n+1} S(k+1) da \leq \int_{0}^{n+1} S(a) da \leq \int_{0}^{n+1} S(k) da $
I don't see how to proceed to get the sum between integrals.
 [/INST] What is $x$ supposed to be in the first line ? What you want to have is
$$ \int_k^{k+1}\frac{a}{x^2+a^2}dx\leqslant\frac{a}{k^2+a^2}\leqslant\int_{k-1}^k\frac{a}{x^2+a^2}dx $$
for all $k\geqslant 1$, which follows from the fact that $x\mapsto\frac{a}{x^2+a^2}$ is decreasing on $[0,+\infty)$. Summing this gives
$$ \int_0^{n+1}\frac{a}{x^2+a^2}dx\leqslant\sum_{k=0}^n\frac{a}{k^2+a^2}\leqslant\frac{1}{a}+\int_0^n\frac{a}{x^2+a^2}dx $$
and taking the limit $n\rightarrow +\infty$,
$$ \int_0^{+\infty}\frac{a}{x^2+a^2}dx\leqslant S(a)\leqslant\frac{1}{a}+\int_0^{+\infty}\frac{a}{x^2+a^2}dx. $$
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