# Trying to figure out a closed form formula

1. Apr 23, 2012

### mesa

Hey guy's, trying to figure out another closed form formula but this time for the sum of 1/squares of the first n consecutive integers.

Or in other words:

1/(1^2) + 1/(2^2) + 1/(3^2) + 1/(4^2) +1/(...=

I tried using the same technique as last time by setting up the formula based on several values for n and then replacing any whole number values with an equation using n but this time the formula gets too complicated since the n value is in the denominator.

Any ideas?

2. Apr 23, 2012

Look for the Riemann zeta function.
en.wikipedia.org/wiki/Riemann_zeta_function

3. Apr 23, 2012

### Whovian

If I remember correctly, I don't know how to prove it, but the result's something crazy like $\displaystyle\dfrac{\sqrt\pi}{6}$

4. Apr 23, 2012

### I like Serena

There are 3 proofs here.

None of them is anything like I'd think up on my own.
Crazy!

5. Apr 23, 2012

### mesa

Hah, neat! Basically it states that for an infinite value for n we get (∏^2)/6 or 1.6449...

I had come up with 1.64... after noticing a pattern with exponents for each group of 10 integers of n.

Basically each group of 10 integers percentage value over the previous would be the value for those ten integers to the power of 1.346756779 then that quantity multiplied by 10. At the point it broke down it left me at 164% or 1.64... with it showing that the value in the hundreths column never exceeding 4. Neat!

6. Apr 23, 2012

### mesa

Nice!

How would you go about tackling this problem?

7. Apr 23, 2012

### I like Serena

(And I was already aware that it is a standard series with a crazy result.)

I'm way too lazy too think up crazy proofs that are already out there.

(But yes, I did think about it for a little while.
I tried to write it as Ʃx^n/n^2 to be evaluated at x=1, then differentiate it, multiply by x, and differentiate again.
But it became too complex to integrate the result, although that may still be possible.)

8. Apr 23, 2012

### A. Bahat

The great thing about the way Euler solved it (with the infinite product) is that it readily generalizes to find the value of the zeta function at any even integer, not just 2. Give it a try!

9. Apr 23, 2012