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
 A: 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
$.
A: You should note that $$\frac{\pi^2}{6}=\sum_{k=1}^\infty\frac{1}{k^2}=\sum_{k=1}^\infty\frac{1}{(2k)^2}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}$$
$$=\frac{1}{4}\sum_{k=1}^\infty\frac{1}{k^2}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{24}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}$$
This shows that 
$$\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$$
Now recognize that 
$$\sum_{k=1}^\infty\frac{1}{\left(k-\frac{1}{2}\right)^2}=\sum_{k=1}^\infty\frac{4}{(2k-1)^2}$$
THis should help lead you to your goal.
A: Simplify the term inside the summation.
$$
\sum \frac{1}{(n-\frac 12)^2} = \sum \frac 4{(2n-1)^2} = 4 \sum \frac 1{(2n-1)^2}
$$
This is just the summation of squares of all odd numbers.
Now, use the fact that:
$$
\frac {\pi^2}6 = \sum \frac 1{k^2} = \left(1 + \frac 1{3^2} + \frac 1{5^2} + \ldots\right) + \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} +  \ldots\right)
$$
And finally, note that:
$$
\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} +  \ldots\right) = \frac 14 \left(\sum \frac 1{k^2}\right) = \frac{\pi^2}{24}
$$
See if you can finish it from here.
A: None of the answers seems to give the final answer, so I'll give it a try here: 
$$\sum_{n=1}^\infty\frac1{\left(n-\frac12\right)^2}=4\sum_{n=1}^\infty\frac1{(2n-1)^2}=4\frac{\pi^2}8=\frac{\pi^2}2$$
and to this we must add the summand corresponding to $\;n=0\;$ , which isn't included there, so we finally get
$$\frac{\pi^2}2+4$$
The above uses
$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\implies$$
$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=\left(\frac16-\frac1{24}\right)\pi^2=\frac{\pi^2}8$$
