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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.

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

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  • $\begingroup$ nice one marco (+1) didn't saw that... $\endgroup$ – tired Jul 19 '16 at 15:05
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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.

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