<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] 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.
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