#### saltydog

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[tex]\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]

First a lemma:

If a polynomial P(x), has non-zero roots [itex]r_i[/itex], and P(0)=1, then:

[tex]P(x)=\left(1-\frac{x}{r_1}\right) \left(1-\frac{x}{r_2}\right) \left(1-\frac{x}{r_3}\right)...\left(1-\frac{x}{r_n}\right)[/tex]

I found that interesting to prove and will leave it for the reader if they wish to do so.

Now consider the polynomial:

[tex]P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .[/tex]

Note that P(0)=1 but we don't know anything about it's roots yet.

Also, consider the power series for Sin(x):

[tex]Sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...[/tex]

Note that:

[tex]xP(x)=Sin(x)[/tex]

Now, since the Sin(x) has roots of 0, and [itex]\pm \pi[/tex], and the 'x' accounts for the zero root on the left, we are left with P(x) containing the remaining roots. Thus P(x) has non-zero roots and we can thus use the lemma above and state:

[tex]

\begin{align*}

1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .&=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)...\\

&=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\left(1-\frac{x^2}{16\pi^2}\right)...

\end{align}

[/tex]

Now I tried multiplying four of those together by hand and with extreme difficulty was able to do so in some manner of order. Apparently Euler was able to do many, many more since he calculated by hand [itex]\zeta(26)[/itex]!

Expanding this product and equating the coefficients to those of P(x) is the key to solving this problem . . .