# Convergence to pi^2/6

1. Feb 24, 2005

### bomba923

How does
infinity
Sum (n^(-2)=(pi^2)/6
n=1

Please tell me if this has been posted before (afraid )
(in that case, i'll see the other post)

Last edited: Feb 24, 2005
2. Feb 24, 2005

### master_coda

Take f(x)=x. Then the Fourier coeffcients of f are $a_n=0$ and $b_n=\frac{2}{n}(-1)^{n+1}$. Parseval's theorem says that:

$$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left(a_k^2+b_k^2)$$

Since the $a_n$ terms are all zero, this reduces to:

$$\frac{1}{\pi}\int_{-\pi}^\pi x^2\:dx=4\sum_{n=1}^\infty\frac{1}{n^2}$$

The integral is easy enough to solve, and the left hand side reduces to $2\pi^2/3$. Dividing both sides by four gives us:

$$\frac{\pi^2}{6}=\sum_{n=1}^\infty\frac{1}{n^2}$$

3. Feb 24, 2005

### bomba923

I see clearly now!--thanks

Last edited: Feb 24, 2005
4. Feb 24, 2005

### dextercioby

And let's not forget Euler's original method.Combining the series he found for $\frac{\sin x}{x}$ and the one from Taylor expansion,he was able to prove it...

Daniel.

5. Feb 24, 2005

### matt grime

And there's another way:

$$\int\int\frac{1}{1-xy}dxdy$$

Evalutate that as x and y both go from 0 to 1. Do it using a substitution, and then do it by replacing the fraction inside with its series expansion and ignore the convergence issues to rearrange sum and integral.

6. Sep 1, 2011

### SpaceDomain

So this is parsevals theorem?

$$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)$$

$$?$$

7. Sep 1, 2011

### Staff: Mentor

$$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty( a_k^2+b_k^2)$$

8. Sep 10, 2012

### mstation

hello! i am new to this forum and it looks like a very nice place!

sorry for my english, i hope everyone can understand it..

sorry for spamming in this thread but it looks like it is the most close to what i need.

i think i understood the answer master_coda gave but i don't understand why he choose f(x)=x..

for instance in my exercise i am asked to verify this equation

$$\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$

what f(x) should i choose for the calculation?

9. Sep 10, 2012

### DonAntonio

I'm answering this, in spite of being an intent of "kidnapping" a thread because

(1) it is, perhaps unwillingly, very close to the OP, and more important

(2) This is a newcomer so he/she doesn't know (but now you do!).

Check the following:

$$\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\Longleftrightarrow \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

DonAntonio