<s>[INST] Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$ Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$  
i know that $ \sum _{k=1} 1/k^2 = \pi^2/6$ how to do prove this
 [/INST] I will show that
$\sum _{n=1}^{\infty} \frac{1}{(n-\frac{1}{2})^2}
=\dfrac{\pi^2}{2}
$.
$$S
=\sum _{n=1}^{\infty} \frac{1}{(n-\frac{1}{2})^2}
=\sum _{n=1}^{\infty} \frac{4}{(2n-1)^2}
=4\sum _{n=1}^{\infty} \frac{1}{(2n-1)^2}
.$$
So this is 4 times
the sum of the odd terms in
$$\sum _{n=1}^{\infty} \frac{1}{n^2}
.$$
To get the sum of the even terms,
$$\sum _{n=1}^{\infty} \frac{1}{(2n)^2}
=\sum _{n=1}^{\infty} \frac{1}{4n^2}
=\frac14\sum _{n=1}^{\infty} \frac{1}{n^2}
=\frac14\frac{\pi^2}{6}
=\frac{\pi^2}{24}
.$$
Combine these two,
using
$\sum _{n=1}^{\infty} \frac{1}{n^2}
=\frac{\pi^2}{6}
$
like this:
$\frac{S}{4}+\frac{\pi^2}{24}
=\frac{\pi^2}{6}
$
or
$\frac{S}{4}
=\frac{\pi^2}{8}
$
or
$S
=\frac{\pi^2}{2}
$.
So your comment appears correct!
As a check,
$\frac{\pi^2}{6}
\approx 1.5
$
and
$S > \frac1{.5^2}
=4
$.
</s>