1. Nov 15, 2005

### johnw188

I saw the other thread, but figured this question was sufficiently distinct to warrent a new thread
I was recently looking at this series
$$$\sum_{n=1}^\infty \frac{1}{n^{2}} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots + \frac{1}{n^{2}} = \frac{\pi^{2}}{6} \approx 1.645$$$
My math teacher gave me the answer of $$\pi^{2}$$/6, and by looking at the sum numerically it seems to come up that way. I'm wondering, though. Where does the $$\pi$$ come from?
I tried to find an expression for the k'th term of the sum, and came up with this
$$\begin{center} \begin{tabular}{| l | c | } \hline k & S_k \\ \hline 1 & 1 \\ \hline 2 & \frac{5}{4} \\ \hline 3 & \frac{49}{36} \\ \hline 4 & \frac{820}{576} \\ \hline 5 & \frac{21076}{14400} \\ \hline 6 & \frac{773136}{518400} \\ \hline \end{tabular} \end{center}$$
Note that the fractions are all left unsimplified. I noticed that all of the denominators were perfect squares:
$$\begin{center} \begin{tabular}{| l | c | } \hline k & S_k \\ \hline 1 & 1^2 \\ \hline 2 & 2^2 \\ \hline 3 & 6^2 \\ \hline 4 & 24^2 \\ \hline 5 & 120^2 \\ \hline 6 & 720^2 \\ \hline \end{tabular} \end{center}$$
As you can see, the denominator of the fraction works out to be k!^2. However, I still can't figure out where the pi comes from, or, for that matter, see any pattern in the numerator. Any ideas?

Last edited: Nov 15, 2005
2. Nov 15, 2005

### hypermorphism

3. Nov 15, 2005

### shmoe

None of the partial sums will have a pi in them, only approximations (the partial sums are all rational).

This has many ways to prove it:

http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf

Depending on what you know, you might find Euler's orignal method (#7 in the above) the easiest to folow. More details on this method can be found in (eq (20) and on):

http://plus.maths.org/issue19/features/infseries/

Though Euler hadn't actually justified his product form for sin(x), it can be done.