 # Proving <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> n = Wesley Hicks 2022-06-02 Answered
Proving
$\sum _{n=1,3,5..}^{\mathrm{\infty }}\frac{4k{\mathrm{sin}}^{2}\left(\frac{n}{k}\right)}{{n}^{2}}=\pi$
Where k any number greater than 0
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Using (in the interval $\left(0,\pi \right)$ )
$\frac{2\pi x-{x}^{2}}{8}=\sum _{n\ge 1}\frac{\mathrm{sin}{\left(n\frac{x}{2}\right)}^{2}}{{n}^{2}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(1\right)$
we have
$\frac{2\pi x-{x}^{2}}{8}=\sum _{n\ge 1}\frac{\mathrm{sin}{\left(\left(2n-1\right)\frac{x}{2}\right)}^{2}}{{\left(2n-1\right)}^{2}}+\frac{1}{4}\sum _{n\ge 1}\frac{\mathrm{sin}{\left(nx\right)}^{2}}{{n}^{2}}$
and using (1) again we get
$\frac{1}{4}\sum _{n\ge 1}\frac{\mathrm{sin}{\left(nx\right)}^{2}}{{n}^{2}}=\frac{\pi x-{x}^{2}}{8}$
so
$\frac{\pi x}{8}=\sum _{n\ge 1}\frac{\mathrm{sin}{\left(\left(2n-1\right)\frac{x}{2}\right)}^{2}}{{\left(2n-1\right)}^{2}}$
now put
$x=\frac{2}{k}$
and we have
$\frac{\pi }{4k}=\sum _{n\ge 1}\frac{\mathrm{sin}{\left(\left(2n-1\right)\frac{1}{k}\right)}^{2}}{{\left(2n-1\right)}^{2}}$
as wanted.