<s>[INST] How Euler get $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$. I saw on wikipedia that he consider $$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$$
The roots are given by $x=\pm n\pi$ and thus (to me)
$$\frac{\sin x}{x}=(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)...$$
How can he get
$$\frac{\sin x}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)\left(1+\frac{x}{3\pi}\right)...\ \ \ ??$$
 [/INST] Euler's proof is a bit adhoc. He uses the fact that for finite polynomials if:
$$(x-r_1)\dots(x-r_n)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$
then $$\dfrac{1}{r_1}+\dots+\dfrac{1}{r_n}=\dfrac{-a_1}{a_0}$$
Then he assumes the result is true for power series, and uses:
$$\dfrac{\sin(\sqrt x)}{\sqrt x}=1-\dfrac{x}{3!}+\dfrac{x^2}{5!}+\dots$$
so $r_k=(k\pi)^2$ and $a_0=1$ and $a_1=\dfrac{-1}{6}$.
Reference
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