Recently in one of my calculus exercise I have made out a (quite novel to me) proof for $\zeta(2)=\frac{\pi^2}{6}$ via the famous infinite product below: $$\sin(x)=x\prod_{i=1}^{\infty}(1-\frac{x^2}{i^2\pi^2})\tag{0}$$ I feel that the proof which I am to give below should count as one proper way to compute $\zeta(2)$ apart from the others (usually very complicated and requiring advanced knowledge) .I think this proof is more accessible to students who are learning very basic calculus and analysis (me, for example). Therefore, I want to improve this proof so that it can really hold water, and as a greenhand at analysis and caculus I need to seek help from this site to improve this proof, which I cannot finish alone. Any help will be appreciated.Thanks in advance.
Here goes the proof:
From $(0)$ we immediately know that $\forall x\ne0$ $$\frac{\sin(x)}{x}=\prod_{i=1}^{\infty}(1-\frac{x^2}{i^2\pi^2})\tag{1}$$ Now let $y=x^2>0$ and rewrite $(1)$ as $$\frac{\sin(\sqrt y)}{\sqrt y}=\prod_{i=1}^{\infty}(1-\frac{y}{i^2\pi^2})\tag{2}$$ For RHS in $(2)$, we may as well regard it as a "polynomial" with an infinite order and also infinitely many terms.Let's just call it $P(y)$. It is then obvious that all the solutions for $P(y)=0$ are as follows: $$y_i=i^2\pi^2$$ where $i\in\mathbb N^+$. Apparently they are distinct.
Now consider a finite polynomial $$P_n(y)=a_0+a_1 y+a_2 y^2+\cdots+a_{n-1}y^{n-1}+a_ny^n$$ Let $P_n(y)=0$, since $y>0$, it is ok to divide both sides by $y^n$, which yields $$a_0 \Bigl(\frac1y\Bigr)^n+a_1\Bigl(\frac1y\Bigr)^{n-1}+\cdots+a_{n-1}\Bigl(\frac1y\Bigr)+a_n=0\tag{3}$$ For each $y_i$ that is a solution to $P_n(y)=0$, $\frac{1}{y_i}$ will also be a solution to $(3)$ if we regard $(3)$ as a polynomial in terms of $\frac1y$. Therefore, by applying the fundamental theorem of algebra to $(3)$ we get $$\sum_{i=1}^{n}\frac{1}{y_i}=-\frac{a_1}{a_0}\tag{4}$$ in which $y_i$s are distinct roots for $P_n(y)=0$.
Then comes the key part, and that's also where I want more clarity on some issues.
Let's return to $P(y)$, then infinite polynomial. By analogue, $(4)$ also holds for $P(y)$ (clarification needed here!!). Therefore, $$\sum_{i=1}^{\infty}\frac{1}{i^2\pi^2}=-\frac{a_1}{a_0}\tag{5}$$ in which $y_i=i^2\pi^2$ are distinct roots for $P(y)=0$ and $a_0$, $a_1$ are respectively the "constant" and "coefficient for $y^1$" in $P(y)$.
To calculate $a_0$ and $a_1$, we must use limits since they are in a limit sense themselves. First, we have $$a_0=\lim_{y\to 0}P(y)=\lim_{y\to 0}\frac{\sin\sqrt y}{\sqrt y}=1\tag{6}$$ Then we go on to calculate $a_1$, recall what we do with a finite polynomial, then by analogue it should be $$a_1=\lim_{y \to 0}\frac{P(y)-a_0}{y}=\lim_{y \to 0}\frac{\frac{\sin\sqrt y}{\sqrt y}-1}{y}=\frac{-\frac16 y+o(y)}{y}=-\frac16\tag{7}$$ (Intuitively I find $(6)$ and $(7)$ acceptable, but I also want some clarification so that they can be convincing.)
Hence, at long last $$\sum_{i=1}^{\infty}\frac{1}{i^2}=-(-\frac{1}{6})\pi^2$$
