# Homework Help: Calculating a Summation

1. Feb 24, 2010

### bcucinel

1. The problem statement, all variables and given/known data

It can be shown that ∑(n=1) to (n=∞) of 1/n² = π²/6

use this fact to show that ∑(n=1) to (n=∞) of 1/(2n-1)² = π²/8

2. Relevant equations

3. The attempt at a solution

2. Feb 24, 2010

### Staff: Mentor

What have you tried? You need to make an attempt at a solution before anyone can give you any help.

3. Feb 24, 2010

### Count Iblis

Hint: It is so easy, you can do it in your head in less than ten seconds.

4. Mar 1, 2010

### bcucinel

My attempt which needs to be shown on paper...

First I wrote out the first few terms of ∑(n=1) to (n=∞) of 1/n². So 1+1/4+1/9+1/16+...+1/n² = π²/6.

I then just tried to simply substituted (2n-1) into the previous summation in place of just n to somehow show that ∑(n=1) to (n=∞) of 1/(2n-1)² = π²/8... but I have no idea how to accurately demonstrate the rest of the proof.
Any additional help would be appreciated.

5. Mar 1, 2010

### Staff: Mentor

What are the first few terms of
$$\sum_{n = 1}^{\infty}\frac{1}{(2n - 1)^2}$$? How does this series relate to the other series?

6. Mar 2, 2010

### bcucinel

When writing out the first few terms of Σ1/(2n-1)², I noticed that this represents the terms when n is some odd number from the first summation, ∑1/n², but I'm stuck as to the proof with setting that equal to 8.

Last edited: Mar 2, 2010
7. Mar 2, 2010

### Staff: Mentor

Show me the first five terms of Σ1/(2n-1)².
Show me the first five terms of Σ1/n².

Pretend for the time being that you don't what the answer is supposed to be. It seems to me that you are too focused on the pi^2/8 result, and not focused enough on how to get there.

8. Mar 2, 2010

### bcucinel

Σ1/(2n-1)²: 1+1/9+1/25+1/49+1/81+...

Σ1/n²: 1+1/4+1/9+1/16+1/25+...

9. Mar 2, 2010

### Staff: Mentor

And which terms are missing from the first series that are in the second series? What is Σ1/n² - Σ1/(2n-1)² ?

10. Mar 2, 2010

### bcucinel

1/4+1/16+1/36+...

Σ1/n² - Σ1/(2n-1)² then equals Σ1/(2n)² from n=1 to infinity

11. Mar 2, 2010

### Staff: Mentor

Now, can you manipulate this series--Σ1/(2n)²--to get to something you know?

12. Mar 2, 2010

### bcucinel

Thank you, I understand all of that perfectly... The issue I am having with the problem, however, is that I don't recall ever being taught in my calculus class exactly how to determine the Sn of a series like Σ1/(2n)²... If there is some technique that I could use please let me know.

13. Mar 2, 2010

### Mathnerdmo

Just simplify 1/(2n)²

14. Mar 2, 2010

### bcucinel

How exactly?

15. Mar 2, 2010

### Staff: Mentor

Write 1/(2n)² in a different way. C'mon, this ain't rocket science...