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 $$\sum_{i=1}^{\infty}\frac{1}{i^2}=-(-\frac{1}{6})\pi^2$$

  • 8
    $\begingroup$ This is basically the same approach as Euler did (see also this answer which also contains 20 different ways to show $\zeta(2)=\pi^2/6$). It is a nice way to motivate the result, but making it a rigorous proof is harder. The biggest problem here is that you cannot treat an infinite polynomial as it was a finite polynomial. $\endgroup$
    – Winther
    Feb 19, 2015 at 18:15
  • 1
    $\begingroup$ @Winther Thank you for so many marvelous proofs that you provided for me! That's too wonderful. But still I don't want to give up my proof. Since $P(y)$ is a convergent infinite polynomial, I think it doesn't matter to write $P(y)$ as $$P(y)=1-\frac16y+\frac{1}{120}y^2+\cdots$$ (in fact I think it is exactly power series as we know it). I believe there must be a fantastic theorem that extends the application of the fundamental theorem of algebra into such convergent infinite polynomial series, and that's what I need to rectify my proof. $\endgroup$
    – Vim
    Feb 20, 2015 at 2:53
  • 3
    $\begingroup$ There isn't. However, if you have proven the product formula and the Taylor series for $\sin(x)/x$ then you could just compare them close to $x=0$ to get a rigorous proof (i.e. consider the limit of $\frac{\sin x - x}{x^3}$ as $x\to 0$ for both expressions to get $-\frac{1}{3!} = -\sum_{n=1}^\infty\frac{1}{n^2\pi^2}$). $\endgroup$
    – Winther
    Feb 20, 2015 at 3:17
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    $\begingroup$ This is exactly the same method that I found too. As Winther has pointed out, asymptotic bounds are good enough to finish it, but the hardest part is actually proving the identity you assumed at the start, rigorously, and that part is not easy at all. It is very tricky to handle all the convergence issues correctly. $\endgroup$
    – user21820
    Apr 29, 2015 at 15:14
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    $\begingroup$ Don't give up! If you're interested, one proof of that uses en.wikipedia.org/wiki/Weierstrass_factorization_theorem. $\endgroup$
    – user21820
    Apr 29, 2015 at 15:29


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