Proof of the value of an infinite sum

  • Thread starter reb659
  • Start date
  • #1
64
0

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?

 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
The sum of 1/(2n)^2 + the sum of 1/(2n+1)^2 = the sum of 1/n^2. Isn't it?
 
  • #3
64
0
...I can't believe I overlooked that. It was sitting in front of me the whole time :(.

Thanks a ton.
 

Related Threads on Proof of the value of an infinite sum

  • Last Post
Replies
4
Views
540
Replies
4
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
985
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
1K
Replies
3
Views
2K
Replies
5
Views
2K
Top