Question about Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$

Question

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.

1. 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?

2. 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.

Answer 1

If you believe Euler's method is totally rigorous, you could also argue in the same way that $$ e^x \frac{\sin x}x = \text{the same product formula you wrote above} $$ because $e^x$ has no complex roots.

Answer 2

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}$.

Answer 3

So you're talking about the Taylor series for $\frac{sin(x)}{x}$, which isn't a polynomial, it's an infinite series which is defined as a limit of polynomials. To see that not all functions representable by Taylor series have zeros, just consider $e^z$. This has a multiplicative inverse $e^{-z}$ and thus is clearly never $0$. It is true that $e^z=1+ z+ \ldots \frac{z^n}{n!} \ldots$, and each of the functions $e_n(z) = 1+z+ \ldots \frac{z^n}{n!}$ has $n$ roots in the complex plane. However, as $n \to \infty$ these roots are "pushed off" the complex plane.

The point is, dealing with Taylor series is subtle, and shouldn't just be thought of as an infinite polynomial, thus we need theorems such as Weierstrass' factorization theorem to deal with them.