# 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.

• No \dfrac in titles please, unless this is absolutely necessary (it was not in the present case). – Did Jul 19 '16 at 14:12

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