# Sum of an infinite alternating series

1. Sep 8, 2006

### DieCommie

I am having trouble finding $$\sum \frac{(-1)^n}{n^2}$$. (from n=1 to n=infinity)

I know that $$\sum \frac{1}{n^2}$$ is $$\frac{\pi^2}{6}$$. But the sum I need is slightly different than that, it is alternating....

Any ideas/help would be appreciated! Thx

2. Sep 8, 2006

The only hint I can give is that the series do converge, since $$lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=lim_{n \rightarrow \infty} \frac{\frac{(-1)^{n+1}}{(n+1)^2}}{\frac{(-1)^n}{n^2}}=lim_{n \rightarrow \infty} \frac{-n^2}{(n+1)^2}=-1 < 1$$, considering d'Alembert's criteria...

3. Sep 8, 2006

### Dragonfall

EDIT: Forget what I said. Mistake.

Last edited: Sep 8, 2006
4. Sep 8, 2006

### shmoe

Hint-find the sum of the even terms of this series.

5. Sep 8, 2006

### DieCommie

and then subtract the odd terms? I must admit I dont know how to sum that series.... When I took calc 2 we only really tested for convergence and divergence.... And Im pouring through my book (stewert 4th ed) and all I see is convergence and divergence, the only sums in my book are pre given (like the one you quoted me on)

Last edited: Sep 8, 2006
6. Sep 8, 2006

### JasonRox

I feel your pain. That book is terrible for learning anything that is formal.

EDIT: Follow shmoe's hint and after that a little algebra should do the rest.

Last edited: Sep 8, 2006
7. Sep 8, 2006

### DieCommie

Ok, I think i got it....

I find the sum of the even terms, which is the sum $$\sum\frac{1}{(2n)^2}$$ which equals $$\frac{\pi^2}{24}$$. I then subtract the even terms from the total and get $$\frac{3\pi^2}{24}$$

Im thinking this is right, but if someone could verify it for me I would appreciate it, and THX so much for the help

EDIT- No, I think I did it backwards. My odd terms are negative, so I need to subtract the odd terms sum from the even terms sum, giving me $$-\frac{3\pi^2}{24}$$

Last edited: Sep 8, 2006
8. Sep 8, 2006

### shmoe

Close! This is what you get when you subract the whole sum from the sum of just the even terms. But what is left in this sum?

You took

$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots$$

and subtracted

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots$$

What terms remain?

9. Sep 8, 2006

### DieCommie

Ok, try again...

what i did was take the even terms and subtract the whole sum... which is not what i wanted.

what I wanted was the even terms minus the odd terms. So I find the odd terms by taking the whole sum minus the even terms.

Then I subtract the odd terms from the even terms and get $$\frac{-\pi^2}{12}$$

God I hope that is right... I have spent alot of time and it is due soon (via internet)

Thx for your help and feed back!! Is $$\frac{-\pi^2}{12}$$ right?!?

10. Sep 8, 2006

### shmoe

That's good!

It's not strictly necessary to find the sum of the odd terms, you can also take 2 times the even terms and then subtract the whole series. Nothing wrong with finding the sum of the odd terms of course, but a (very slightly) different view might be usefull to see.