Prove-that-1-1-2-2-1-3-2-1-4-2-pi-2-6-

Question Number 44301 by naka3546 last updated on 26/Sep/18

P r o v e t h a t

1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \dots = \frac{π^{2}}{6}

Commented by abdo.msup.com last updated on 26/Sep/18

l e t c o n s i d e r f (x) =∣ x ∣ (2 π p e r i o d i c e v e n)

l e t d e v e l o p p f a t f o u r i e r s e r i e

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x)

a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x

= \frac{2}{2 π} \int_{- π}^{π} ∣ x ∣ c o s (n x) d x = \frac{2}{π} \int_{0}^{π} x c o s (n x) d x

\Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x c o s (n x) d x

= {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x

= - \frac{1}{n} \int_{0}^{π} s i n (n x) d x

= - \frac{1}{n} {[- \frac{1}{n} c o s (n x)]}_{0}^{π}

= \frac{1}{n^{2}} {{(- 1)}^{n} - 1} \Rightarrow \frac{π}{2} a_{n} = \frac{{(- 1)}^{n} - 1}{n^{2}}

a_{n} = \frac{2}{π} \frac{{(- 1)}^{n} - 1}{n^{2}} \Rightarrow

\frac{π}{2} a_{0} = \int_{0}^{π} x d x = {[\frac{x^{2}}{2}]}_{0}^{π} = \frac{π^{2}}{2} \Rightarrow \frac{a_{0}}{2} = \frac{π}{2}

∣ x ∣ = \frac{π}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} - 1}{n^{2}} c o s (n x)

= \frac{π}{2} - \frac{4}{π} \sum_{n = 0}^{\infty} \frac{c o s (2 n + 1) x}{{(2 n + 1)}^{2}}

x = 0 \Rightarrow \frac{π}{2} - \frac{4}{π} \sum_{n = 0} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

\frac{4}{π} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π}{2} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}

b u t \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{p = 1}^{\infty} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}}

= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \frac{π^{2}}{8} \Rightarrow

\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4 π^{2}}{24}

= \frac{π^{2}}{6} .