Inner Product Examples, what is the points? Example:
For $ -\pi<x<\pi$, 
$$x =-2  \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$$  
and
$$x^3 =-2  \sum_{n=1}^{\infty} \left( \frac{\pi^2}{n}-\frac{6}{n^3} \right)(-1)^n \sin(nx)$$
by using inner products of these two functions, the value of
$$\sum_{n=1}^{\infty} \frac{1}{n^4}$$
is equal to $\frac {\pi^4}{90} $ .

My question is what is the point in this example that the author gives
  the solution without any detail? any users can tips me how this value
  is reached?

 A: HINT:
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4} $$
FURTHER EXPLANATION:
You have two functions $f(x)=x$ and $g(x)=x^3$. The coefficients of the expansion of $f$ in the basis $\sin(nx)$ are of the form $$a_n=-2\times \frac{(-1)^n}{n}$$
and the coefficients for $g(x)$ are: $$b_n=-2\times\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n$$
Then the inner product is given by: 
$$\frac 1\pi\int_{-\pi}^{\pi}f(x)g(x)=\sum_{n=1}^\infty a_n b_n$$
Replacing $f,g,a_n$ and $b_n$, using the fact that $\sum 1/n^2=\pi^2/6$ and manipulating, gives you the answer.
SOLUTION:

$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \left(-2\times \frac{(-1)^n}{n}\right)\times(-2)\left(\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n\right) $$
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4} $$
$$\frac{2\pi^5}{5\pi}=4\pi^2\times\frac{\pi^2}6-24\sum_{n=1}^\infty\frac 1{n^4}$$
$$24\sum_{n=1}^\infty\frac 1{n^4}=\pi^4\left( \frac 23 - \frac 25\right) $$
$$\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{90}$$

