Fault in proof of $\zeta(2) = \frac{\pi^2}{6}$

Asked Dec 20 '15 at 15:12

Active Dec 20 '15 at 15:20

Viewed 144 times

Consider the proof of: $$\zeta(2) = \frac{\pi^{2}}{6}$$

So the proof assume that (because of Euler decomposition) $$\frac{\sin(x)}{x} = \prod_{n > 0}\left(1 - \frac{x^{2}}{(n\pi)^{2}}\right)$$

Why should we consider the $\frac{\sin(x)}{x}$, that $\sin(x)$ has the same decomposition?

Please tell me where's my fault.

edited Dec 20 '15 at 15:20

asked Dec 20 '15 at 15:12

openspace

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http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2?lq=1 – Jack D'Aurizio Dec 20 '15 at 15:19
The infinite product is equal to $1$ if $x = 0$, but $\sin 0 = 0$. – Andrew D. Hwang Dec 20 '15 at 15:26
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Why should $\sin x$ have the same decomposition? – egreg Dec 20 '15 at 15:30
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@AndrewD.Hwang but $x=0$, then $\frac{\sin x}{x}$ does not exist and its limit is equal to $1$. – Kamil Jarosz Dec 20 '15 at 15:42
@egreg yes, that's my fault. And I forgot about Teylor's sum. – openspace Dec 20 '15 at 15:43
@kamil09875: If you're going to treat $\frac{\sin x}{x}$ as undefined at $0$ rather than considering the analytic (convergent power series) extension, then the OP's equation doesn't make sense at all. I was merely pointing out why the infinite product cannot be equal to $\sin x$. – Andrew D. Hwang Dec 20 '15 at 17:12

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