# Infinite series wizardry

1. Sep 1, 2011

### Screwdriver

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

Check out matt grime's post in this thread (it's the last one):

How exactly did he know that the sum could be represented as that double integral? Also, is there a method of converting sums like that to integrals (double or otherwise) for summands other than $n^{-2}$ such as $n^{-7}$ or something?

2. Relevant equations

$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy=\sum_{n=1}^{\infty }\frac{1}{n^2}$$

3. The attempt at a solution

Come to think of it, I don't even really see how that helps you, because the series expansion for the $y$ integral (after computing the $x$ integral) is the derivative of the series you're trying to find, so integrating it just brings you back to where you started. The only thing I was able to note was that:

$$\sum_{n=0}^{\infty} (xy)^n=\frac{1}{1-xy}$$

Last edited by a moderator: Apr 26, 2017
2. Sep 1, 2011

### micromass

$$\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum{\frac{1}{n^2}}$$

Well, you're on the right way. Just notice that

$$\frac{1}{1-xy}=\sum{(xy)^{n-1}}$$

So by monotone convergence, we can switch the limit and the integral: So

$$\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum\int_0^1\int_0^1 (xy)^{n-1}dxdy$$

So you only need to calculate $\int_0^1\int_0^1 (xy)^{n-1}dxdy$ now...

3. Sep 1, 2011

### Screwdriver

Well what I'm really asking is how to convert the sum to an integral. Like If I just saw $\sum_{n=1}^{\infty }\frac{1}{n^2}$ somewhere, I would start with this:

$$\sum_{n=1}^{\infty} (xy)^{n-1}=\frac{1}{1-xy}$$

But how did I know to start with that? It appears to have no relation to $\sum_{n=1}^{\infty }\frac{1}{n^2}$ ... and then all of a sudden there's a double integral with non-infinite limits.

4. Sep 1, 2011

### lineintegral1

Recall the (real) geometric series,

$\frac {1}{1-x}=\sum_{n=0}^{\infty}x^n$

This is true provided, of course, that $|x|<1$.

Anytime you see a rational expression of this form, a light should go off in your head regarding the geometric series. It should be easy to see now how to prove the equivalence of your double integral and $\zeta(2)$.

Not sure if this clarifies anything; let us know if you have more questions.

5. Sep 1, 2011

### Screwdriver

It's not showing that the integral and sum are equivalent that's my problem, it's coming up with the integral in the first place . Let me try to rephrase this:

Given a sum, I want to know how to write down a computable integral whose value is the value of that sum.

6. Sep 3, 2011

### Screwdriver

First and last bump for this thread.