1. ## Basel problem application

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

2. Originally Posted by Lord Darkin
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
Start by noting that $\displaystyle \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}$.

3. Okay. I'm not sure how you got that, though.

Also, I'm not sure how to proceed even with that.

Am I supposed to use a taylor series anywhere during this problem? I've looked at Wiki's page and it says that sinx/x was a proof used to find pi^2/6

4. Originally Posted by Lord Darkin
Okay. I'm not sure how you got that, though.

Also, I'm not sure how to proceed even with that.

Am I supposed to use a taylor series anywhere during this problem? I've looked at Wiki's page and it says that sinx/x was a proof used to find pi^2/6
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.

5. Originally Posted by TheEmptySet

$\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2}+...=\left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{2^2}+\frac{1}{4^2}+...\right)$

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 ... ?

6. Originally Posted by Lord Darkin
$\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2} + ... = \left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2$

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 ... ?
Please go back and read the replies you got again. In particular, my first reply shows exactly what to do.

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

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

$\displaystyle ( \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1} {4^2}+\frac{1}{5^2} + ... ) - ( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1} {8^2}+\frac{1}{10^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}$

$\displaystyle ( \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1} {7^2}+\frac{1}{9^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}$

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.

8. Originally Posted by Lord Darkin
$\displaystyle \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}$

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

[snip]
Therefore:

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

and I hope it's clear what happens next.