Thread: Partial Sums

Given that the summation from 1 to infinity of 1/n^2 = pi^2/6, evaluate
a) the summation from 0 to infinity of 1/(2n+1)^2
b) the summation from 0 to infinity of 1/(4n+1)^2

$\displaystyle \sum_{n=1}^{+\infty}\frac{1}{n^2}\;=\;\frac{\pi^2} {6}$

$\displaystyle \sum_{n=1}^{+\infty}\frac{1}{(2n)^2}\;=\;\sum_{n=1 }^{+\infty}\frac{1}{4n^2}$

$\displaystyle \sum_{n=1}^{+\infty}\frac{1}{(2n)^2}\;=\;\frac{1}{ 4}\sum_{n=1}^{+\infty}\frac{1}{n^2}\;=\;\frac{\pi^ 2}{24}$

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

$\displaystyle \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}\;=\;\frac{\ pi^2}{6}-\frac{\pi^2}{24}=\;\frac{\pi^2}{8}$

I got beaten but I'll post anyway,

$\displaystyle \sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}=\sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}+\frac{1}{4}\sum\limits_{n=1}^{\ infty }{\frac{1}{n^{2}}},$ hence your sum equals $\displaystyle \left( \sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}+\frac{1}{4}\sum\limits_{n= 1}^{\infty }{\frac{1}{n^{2}}} \right)-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}$ and this is $\displaystyle \sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}=\frac{3}{4}\sum\limits_{n=1}^{\ infty }{\frac{1}{n^{2}}}=\frac{\pi ^{2}}{8}.$

You can do the second one in the same fashion.