By using Wallis’s product, prove $\zeta(2)=\frac{\pi^2}{6}$ $\sin x = x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\cdots$  
This equal is used in both the solution of  Basel’s problem and the proof of Wallis’s product.
So I wonder if I can prove one by using the other.
Help me solve this problem.
Or if you are sure this assumption is wrong, tell me the reason.   
I’m sorry if I put wrong tugs.
 A: Given
$$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2 \pi^2}\right) \tag{Weierstrass}$$
we have, by applying $\frac{d^2}{dx^2}\log(\cdot)$ to both sides,
$$ \frac{1}{x^2}-\frac{1}{\sin^2 x} = -2\sum_{n\geq 1}\frac{x^2+n^2\pi^2}{(\pi^2 n^2-x^2)^2}\tag{SecondLogDerivative}$$
hence:
$$ \zeta(2) = \frac{\pi^2}{2}\lim_{x\to 0^+}\left(\frac{1}{\sin^2 x}-\frac{1}{x^2}\right)\stackrel{\text{De l'Hopital}}{=} \frac{\pi^2}{6}.\tag{BaselProblem} $$

Here it is another crazy approach, not really related to Wallis product, but interesting in itself (at least I hope). For any $s>1$ we have
$$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s}=\left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}\stackrel{\mathcal{L}^{-1}}{=}\frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt $$
where the RHS is converging for any $s>0$, providing an analytic continuation of the LHS over such region. By applying integration by parts twice, we get the following integral representation for the $\zeta$ function over the region $s>-2$:
$$ \zeta(s)=\frac{1}{\Gamma(s+2)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{t^{s+1}e^t(e^t-1)}{(e^t+1)^3}\,dt \qquad(\text{IntegralRepFor}\zeta)$$
and due to the reflection formula $\frac{\zeta(1-s)}{\zeta(s)}=\frac{2\,\Gamma(s)}{(2\pi)^s}\,\cos\left(\tfrac{\pi s}{2}\right) $ we have
$$ \zeta(s)=\frac{(2\pi)^s}{2\,\Gamma(3-s)\,\Gamma(s)\cos\frac{\pi s}{2}}\left(1-2^s\right)^{-1}\int_{0}^{+\infty}\frac{t^{2-s}e^t(e^t-1)}{(e^t+1)^3}\,dt \qquad(\zeta- \text{Reflection})$$
for any $s<3$. By evaluating the previous line at $s=2$ and by enforcing the substitution $t=\log u$ we get:
$$ \zeta(2)=\frac{2\pi^2}{3}\int_{1}^{+\infty}\frac{u-1}{(u+1)^3}\,du\stackrel{u\mapsto\frac{1}{1-v}}{=}\frac{2\pi^2}{3}\int_{0}^{1}\frac{v}{(2-v)^3}\,dv=\frac{2\pi^2}{3}\cdot\frac{1}{4}=\color{red}{\frac{\pi^2}{6}}.\qquad(\text{MiraclesHappen})$$

Another twist of classical approaches.
$$\zeta(2)=\lim_{n\to +\infty}[z^n]\frac{\text{Li}_2(z)}{1-z}=\lim_{n\to +\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\text{Li}_2(e^{i\theta})}{e^{ni\theta}(1-e^{i\theta})}\,d\theta$$
equals
$$\lim_{n\to +\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\text{Li}_2(e^{i\theta})(e^{-ni\theta}-e^{-(n+1)i\theta})}{2-2\cos\theta}\,d\theta $$
or
$$ \lim_{M\to +\infty}\sum_{n=1}^{M}\frac{\left(1-\frac{n}{M}\right)}{n^2}=\lim_{m\to +\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}F_M(x)\sum_{n\geq 1}\frac{\cos(n\theta)}{n^2}\,d\theta $$
where $F_M$ is Fejér kernel and the involved Fourier series is a $2\pi$-periodic, even, piecewise-parabolic function with mean zero and second derivative equal to $\frac{1}{2}$ over $\mathbb{R}\setminus \pi\mathbb{Z}$. In particular the involved Fourier series, $\text{Re}\,\text{Li}_2(e^{i\theta})$, equals $\frac{1}{4}(\theta-\pi)^2+C$ for $\theta\in(0,2\pi)$ and $C$ has to be $-\frac{\pi^2}{12}$ in order to ensure that such function has mean zero. Then we have
$$ \zeta(2)=\lim_{\theta\to 0^+}\text{Re}\,\text{Li}_2(e^{i\theta})=\lim_{\theta\to 0^+}\frac{1}{4}(\theta-\pi)^2-\frac{\pi^2}{12}=\frac{\pi^2}{6}.$$
A: Define $I_{n}$ as $$I_{n}=\int_{0}^{\frac{\pi}{2}} \cos^{2n} x dx$$
and $J_{n}$ as $$J_{n}=\int_{0}^{\frac{\pi}{2}} x^{2}\cos^{2n} x dx$$
Use partial integration, and we get
$$I_n=n(2n-1)J_{n-1}-2n^2J_{n}\cdots①$$
Here, Wallis’s product is
 $$ I_n=\frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}$$
$$⇔I_n= \frac{(2n)!}{4^n n!^2}\frac{\pi}{2}$$
So,
$$①⇔\frac{(2n)!}{4^n n!^2}\frac{\pi}{2}=n(2n-1)J_{n-1}-2n^2J_{n}\ $$
$$⇔\frac{\pi}{4}\frac{1}{n^2}=\frac{4^{n-1}(n-1)!^2}{(2n-2)!}J_{n-1}-\frac{4^{n}(n)!^2}{(2n)!}J_{n}$$
Let $K_n$ be $\frac{4^{n}(n)!^2}{(2n)!}J_{n} $
The sum from $n=1$ to $N$ is
$$\frac{\pi}{4}\sum_{n=1}^{N} \frac{1}{n^2}=K_0-K_N$$
Here, $K_0=J_0=\frac{\pi^3}{24}$
So, 
 $$\frac{4^{N}(N)!^2}{(2N)!}J_{N}=\frac{\pi}{4} \left(\frac{\pi^2}{6}-\sum_{n=1}^{N} \frac{1}{n^2} \right)\cdots②$$  
$0<x<\frac{\pi}{2}$⇒$0<x<\frac{\pi}{2}\sin x$
So, 
$$0<J_{n}<\frac{\pi^2}{4}\int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^{2n} x= \frac{\pi^2}{4}\left(I_n-I_{n+1}\right) $$
$$⇔0<\frac{4^{N}(N)!^2}{(2N)!} J_N<\frac{\pi^3}{16(N+1)}$$
By the squeeze theorem, we can get
$$\lim_{n→∞} \frac{4^{N}(N)!^2}{(2N)!} J_N=0$$
$$∴ \sum_{n=1}^{N} \frac{1}{n^2}=\frac{\pi^2}{6} (∵②)$$
A: You get the $\,$ Wallis product $\,$, if you set $\,\displaystyle x=\frac{\pi}{2}\,$ .
Basel problem : 
The coefficient for $\,z\,$ of the infinite product $\,(1+a_1 z)(1+a_2 z)...\,$ is $\,a_1+a_2+...\,$ . 
Therefore with $z:=-x^2$ and using the series of sine we get $\,\displaystyle\frac{1}{3!}=\frac{1}{(1\cdot \pi)^2}+\frac{1}{(2\cdot \pi)^2}+...\,$ .

Both calculations are based on partial fraction expansion or equivalent on the sine product :
$($ Wallis product $\displaystyle )^2 \,\,= \left(\prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)^{-1}|_{x=\frac{1}{2}}\right)^2=\left(\frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}}\right)^2=\frac{\pi^2}{4}$
$\displaystyle \frac{\pi^2}{6}=\frac{1-\pi x \cot(\pi x)}{2 x^2}|_{x\to 0}=\sum\limits_{k=0}^\infty \frac{1}{k^2-x^2}|_{x=0}=\zeta(2) $
Both calculations can be put together to an equation chain. The core thing is:
$\displaystyle \frac{2}{3}\prod\limits_{k=1}^\infty \left(1-\left(\frac{x+\frac{1}{2}}{k}\right)^2\right)^{-2} \bigg|_{x=0} =\frac{\pi^2}{6} = -\frac{1}{2}\frac{d^2}{dx^2}\ln \prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)\bigg|_{x=0} $
