Use the Table of Fourier Series with suitable values of $x$ to evaluate the following series $\sum \frac{1}{n^2}$

Question

I have the answer but not the procedure.

Using the Table of Fourier Series: $ \frac{L^2}{3} + \frac{4L^2}{\pi^2}\cdot∑\frac{(-1)^n}{n^2}\cdot\cos(\frac{n\cdot\pi\cdot x}{L}) ;\\ f(x) = x^2 ; -L<x<L$

Using this show that the answer is $ ∑=\frac{\pi^2}{6}$

Answer 1

Parseval identity :$$\frac{a^2_0}{2}+\sum\limits_{n = 1 }^\infty a^2_n+b^2n = \frac{1}{{2\pi }}\int_{ - \pi }^\pi | f(x){|^2}{\mkern 1mu} dx$$ you can write Fourier series for $f(x)=x ; x \in (-\pi,\pi],T=2\pi$

$$\begin{array}{*{20}{l}} {f(x)}&{ = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left[ {{a_n}\cos \left( {nx} \right) + {b_n}\sin \left( {nx} \right)} \right]} }\\ {}&{ = \sum\limits_{n = 1}^\infty {2\frac{{{{( - 1)}^{n + 1}}}}{n}} \sin (nx),\quad .} \end{array}$$ now use identity $$\frac{a^2_0}{2}+\sum\limits_{n = 1 }^\infty a^2_n+b^2n = \frac{1}{{2\pi }}\int_{ - \pi }^\pi | f(x){|^2}{\mkern 1mu} dx\\ \frac{0}{2}+\sum\limits_{n = 1 }^\infty 4(\dfrac{(-1)^n}{n})^2 = \frac{1}{{2\pi }}\int_{ - \pi }^\pi x{^2}{\mkern 1mu} dx$$ so $$\sum\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$

Answer 2

HINT:put $x=\frac L2$ $$\frac{L^2}{3} + \frac{4L^2}{\pi^2}\cdot∑\frac{(-1)^n}{n^2}\cdot\cos(\frac{n\cdot\pi\cdot x}{L}) ;\\ f(x) = x^2 ; -L<x<L \to $$ so $$\frac{L^2}{3} + \frac{4L^2}{\pi^2}\cdot∑\frac{(-1)^n}{n^2}\cdot\cos(\frac{n\cdot\pi}{2})=\frac{L^2}{4}$$ can you go on ? $$\frac{1}{3} + \frac{4}{\pi^2}\cdot∑\frac{(-1)^n}{n^2}\cdot\cos(\frac{n\cdot\pi}{2})=\frac{1}{4}\\ \frac{4}{\pi^2}\cdot∑\frac{(-1)^n}{n^2}\cdot\cos(\frac{n\cdot\pi}{2})=\frac{-1}{12}$$\simplify $$∑\frac{(-1)^n}{n^2}=\frac{-\pi^2}{48}\\\frac{-1}{1^2}0+\frac{(-1)^2}{2^2}(-1)+\frac{(-1)^3}{3^2}0+\frac{(-1)^4}{4^2}(+1)+...=\frac{-\pi^2}{48}$$ multiply by(-4) $$1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}-\frac{1}{25}+...=\frac{+\pi^2}{12}\\$$ Suppose $A=∑\frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...$ now we can rewrite $$1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}-\frac{1}{25}+...$$ as $$1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}-\frac{1}{25}+...= \\1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...-2(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...)=\frac{\pi^2}{12}\\\to 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...-2\frac 14(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{25}+\frac{1}{36}+...)=\frac{\pi^2}{12}\\ \underbrace{1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...}_{A}-2\frac 14\underbrace{(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{25}+\frac{1}{36}+...)}_{A} \\\to A-\frac12A=\frac{\pi^2}{12}\\A=\frac{\pi^2}{6}$$