Prove that $\displaystyle \frac{\pi^2}{8} = \sum \frac{1}{(2n-1)^2}$
How do I go about starting this, knowing basel's problem?
Basel problem - Wikipedia, the free encyclopedia
Prove that $\displaystyle \frac{\pi^2}{8} = \sum \frac{1}{(2n-1)^2}$
How do I go about starting this, knowing basel's problem?
Basel problem - Wikipedia, the free encyclopedia
As Mr. Fantastic Suggested he is breaking the sum up into the even and odd parts.
for example
$\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2}+...=\left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{2^2}+\frac{1}{4^2}+...\right)$
As another hint factor a $\displaystyle 2^2$ out of the even terms on the right hand side and compare the right and left hand sides of the equality.
$\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2} + ... = \left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2$
Then that gives me
$\displaystyle \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1} {8^2}+\frac{1}{10^2}+...= +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2$
So ... I have now factored out the 2 squared. But that just leaves me with the same thing I had ... ?
$\displaystyle \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}$
$\displaystyle \sum_{n=1}^{+ \infty} \frac{1}{n^2} - \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}$
$\displaystyle ( \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2} + ... ) - ( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1} {8^2}+\frac{1}{10^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}$
$\displaystyle ( \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1} {7^2}+\frac{1}{9^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}$
Sorry about this I tried looking at Mr fantastic posts but I still cannot deciper it. I know that I'm close but I can't get it.