Sum of an infinite alternating series

[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

 

[radou]

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...

 

[Dragonfall]

EDIT: Forget what I said. Mistake.

 

[shmoe]

I know that \sum \frac{1}{n^2} is \frac{\pi^2}{6}.

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

 

[DieCommie]

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

and then subtract the odd terms? I must admit I don't know how to sum that series... When I took calc 2 we only really tested for convergence and divergence... And I am 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)

 

[JasonRox]

and then subtract the odd terms? I must admit I don't know how to sum that series... When I took calc 2 we only really tested for convergence and divergence... And I am 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)

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.

 

[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}

 

[shmoe]

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}

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?

 

[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 a lot of time and it is due soon (via internet)

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

 

[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.