# summation related to reciprocals

I know that $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$ in a test this was given and we were asked to find $\sum \frac{1}{(2n+ 1)^2}$ starting with $n=0...\infty$. Now i am grade $11$ student and have been here for a while so i know its proved by Euler if I am not mistaken . But as a student of such a low grade I would like to prove the summation using some higher algebra and basic calculus . Hope you guys help.

• Add $\sum\frac1{(2n)^2}=\frac14\sum\frac1{n^2}$ to the sum you are looking for and marvel that the result is $\sum\frac1{n^2}$ itself. – Did Feb 21 '16 at 12:33
• So what woukd the resultant answer be ? – Archis Welankar Feb 21 '16 at 12:34
• Hmmm... This is rather direct from my comment, no? – Did Feb 21 '16 at 12:35
• I think it would be $Π^2/8$ – Archis Welankar Feb 21 '16 at 12:38
• So can we also do it by limits or not – Archis Welankar Feb 21 '16 at 12:40

Divide the sum in odd and even denominators. This can be done since the series if absolutely convergent and thus we can change the order of summation without affecting the sum, and use what the comment below your question implies:

$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac{\pi^2}{24}+\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

Now substract and that's all, but observe that if your series begins with $\;n=1\;$ then you have one summand more than the above series.

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{n^2}$$

This is true because the left sum (we want to know it, call it $S$) sums over the reciprocals of the odd squares and the right sum does the same for all even squares.

We can remove $\frac{1}{4}$ from the right hand sum, as $\frac{1}{(2n)^2} = \frac{1}{4} \frac{1}{n^2}$ for all $n$.

So $S + \frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{6}$, hence $S$ = $\frac{3}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}$.