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## Homework Statement

On an exam we were given that the [tex]\Sigma[/tex] 1/n^2 was (pi^2)/6. We were asked to prove that the value of [tex]\Sigma[/tex] ((-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.

## Homework Equations

Basic properties of series.

[tex]\Sigma[/tex] 1/n^2 = (pi^2)/6

**3. The Attempt at a Solution**

**I rewrote the series as:**

[tex]\Sigma[/tex] 1/(2n+1)^2 - [tex]\Sigma[/tex]1/(2n)^2

From the given information I know that [tex]\Sigma[/tex] 1/(2n)^2 = (pi^2)/24.

The problem is that fact that I don't know how to find [tex]\Sigma[/tex] 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 [tex]\Sigma[/tex] 1/(2n+1)^2?[tex]\Sigma[/tex] 1/(2n+1)^2 - [tex]\Sigma[/tex]1/(2n)^2

From the given information I know that [tex]\Sigma[/tex] 1/(2n)^2 = (pi^2)/24.

The problem is that fact that I don't know how to find [tex]\Sigma[/tex] 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 [tex]\Sigma[/tex] 1/(2n+1)^2?

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