# Proof of the value of an infinite sum

1. Sep 11, 2008

### reb659

1. The problem statement, all variables and given/known data
On an exam we were given that the $$\Sigma$$ 1/n^2 was (pi^2)/6. We were asked to prove that the value of $$\Sigma$$ ((-1)^(n+1))/(n^2)) = (pi^2)/12.

I'm sorry about the lack of latex; this is the first time I've ever *tried* to use it.

2. Relevant equations
Basic properties of series.
$$\Sigma$$ 1/n^2 = (pi^2)/6

3. The attempt at a solution
I rewrote the series as:
$$\Sigma$$ 1/(2n+1)^2 - $$\Sigma$$1/(2n)^2
From the given information I know that $$\Sigma$$ 1/(2n)^2 = (pi^2)/24.
The problem is that fact that I don't know how to find $$\Sigma$$ 1/(2n+1)^2.
I used the integral test to approximate it, and I got 5/4, which is extremely close to the value it should be, which is (pi^2)/8, but I don't think this counts as "proving" it.

Can anyone think of an alternative way to go about proving it, or finding the exact value of $$\Sigma$$ 1/(2n+1)^2?

Last edited: Sep 11, 2008
2. Sep 11, 2008

### Dick

The sum of 1/(2n)^2 + the sum of 1/(2n+1)^2 = the sum of 1/n^2. Isn't it?

3. Sep 11, 2008

### reb659

...I can't believe I overlooked that. It was sitting in front of me the whole time :(.

Thanks a ton.