Question about Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$ For a freshman calculus project, I used Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, and noted from Wikipedia's explanation that the infinite product representation of $\frac{\sin x}x=\prod_{n=1}^\infty(1-\frac{x^2}{n^2\pi^2})$ is unjustified without Weierstrass' factorization theorem. I'm finding it very difficult to follow the article about Weierstrass' theorem.


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*Can someone explain to me what's unjustified about Euler's infinite product representation? Since $\frac{\sin(x)}x$ has a Taylor polynomial representation, and I think all polynomials have roots (in the set of complex numbers), shouldn't it also have a infinite product of roots representation?

*Can someone explain to me what Weierstrass' theorem does to justify Euler's representation, and if it's within the ability of a freshman calculus student, can someone show me a proof that is more accessible that the ones I've found by Googling?
Thanks for your time. This is a very interesting problem and very different from the ones I'm used to doing in my calculus class.

There is always the option of evaluating the infinite product directly. But I don't know how much you know about the gamma function.
Euler's limit definition of the gamma function is $ \displaystyle\Gamma(z) = \lim_{m \to \infty} \frac{m! \ m^{z}}{z(z+1)\ldots (z+m)}$.
Then $$ \displaystyle \Gamma(1+z) = z \Gamma(z) = \lim_{m \to \infty} \frac{m! \ m^{z}}{(z+1)\cdots (z+m)} = \lim_{m \to \infty} m^{z} \prod_{k=1}^{m} \left(1+\frac{z}{k} \right)^{-1} $$
And $$ \Gamma(1-z) = \lim_{m \to \infty} m^{-z} \prod^{m}_{k=1} \left( 1- \frac{z}{k} \right)^{-1}$$
So $$ z \Gamma(z) \Gamma(1-z) = \lim_{m \to \infty} \prod_{k=1}^{m} \left( 1 - \frac{z^{2}}{k^{2}} \right)^{-1} =  \prod_{k=1}^{\infty} \left( 1 - \frac{z^{2}}{k^{2}} \right)^{-1}$$
Or using the reflection formula for the gamma function,
$$\frac{ \pi z}{\sin \pi z} = \prod_{k=1}^{\infty} \left( 1 - \frac{z^{2}}{k^{2}} \right)^{-1} \implies \frac{\sin \pi z}{\pi z} = \prod_{k=1}^{\infty} \left( 1 - \frac{z^{2}}{k^{2}} \right)$$
Finally just replace $z$ with $ \displaystyle \frac{z}{\pi}$.
