Proof that $\sum\limits_{k=1}^\infty\frac{2k}{(k^2+c^2)^2}\gt\frac{2}{2c^2+1}$ I tried to prove the following inequality which gives a lower bound to the Mathieu sum:
$$S=\sum_{k=1}^\infty\dfrac{2k}{(k^2+c^2)^2}$$
where $c\neq0$.
The Mathieu inequality states: $S\lt\dfrac{1}{c^2}$
The following inequality holds: 
$$S\gt\dfrac{1}{c^2+\dfrac{1}{2}}$$
I tried to expand $S$ and I found an expression for $S$ very difficult to manage, so it seems very hard to follow this way to prove it. Is there a better method to prove it? Thanks.
 A: You may observe that $$\frac{2k}{\left(k^{2}+c^{2}\right)^{2}+\frac{1}{4}+c^{2}}<\frac{2k}{\left(k^{2}+c^{2}\right)^{2}}\tag{1}
 $$ and $$\frac{2k}{\left(k^{2}+c^{2}\right)^{2}+\frac{1}{4}+c^{2}}=\frac{1}{\left(k-\frac{1}{2}\right)^{2}+\frac{1}{4}+c^{2}}-\frac{1}{\left(k+\frac{1}{2}\right)^{2}+\frac{1}{4}+c^{2}}
 $$ hence, if we take the sum in $(1)$ we get 

$$\frac{1}{c^{2}+\frac{1}{2}}<\sum_{k\geq1}\frac{2k}{\left(k^{2}+c^{2}\right)^{2}}.$$

A similar trick works for the other inequality.
A: It is a bit an overkill, but since for any $a,b>0$ we have:
$$ \int_{0}^{+\infty}x\sin(ax)e^{-bx}\,dx = \frac{2ab}{(a^2+b^2)^2} \tag{1}$$
it happens that:
$$ \sum_{k\geq 1}\frac{2k}{(k^2+c^2)^2}=\int_{0}^{+\infty}\sum_{k=1}^{+\infty}\frac{x\sin(cx)}{c}e^{-kx}\,dx = \frac{1}{c}\int_{0}^{+\infty}\frac{x\sin(cx)}{e^x-1}\,dx\tag{2} $$
and if $|c|\leq 1$ we have:
$$ \sum_{k\geq 1}\frac{2k}{(k^2+c^2)^2}= 2\zeta(3)-4c^2\zeta(5)+6c^4\zeta(7)-8c^6\zeta(9)+\ldots\tag{3}$$
On the other hand, by partial fraction decomposition and the identity $\sum_{n\geq 0}\frac{1}{(n+a)^2}=\psi'(a)$  we have:
$$ \sum_{k\geq 1}\frac{2k}{(k^2+c^2)^2}=\frac{\text{Im}\left(\psi'(1-ic)\right)}{c}\tag{4}$$
so for large values of $c$ we may use Stirling's approximation for $\log\Gamma$ or the Euler-Maclaurin summation formula: notice that
$$ \int_{1}^{+\infty}\frac{2x\,dx}{(x^2+c)^2}=\frac{2}{2c^2+2}.\tag{5}$$
The connection between my approach and Marco's one is very nice: Stirling's inequality is a consequence of creative telescoping.
