prove-that-r-1-1-r-2-pi-2-6-

Question Number 76367 by Rio Michael last updated on 26/Dec/19

p r o v e t h a t \underset{r = 1}{\sum^{\infty}} \frac{1}{r^{2}} = \frac{π^{2}}{6}

Commented by mathmax by abdo last updated on 26/Dec/19

n o t c o r r e c t

Commented by mathmax by abdo last updated on 26/Dec/19

i t h i n k \sum_{r = 1}^{\infty} \frac{1}{r^{2}}

Commented by Rio Michael last updated on 26/Dec/19

t h a n k s i^{'} l l c o r r e c t t h a t

Commented by mathmax by abdo last updated on 28/Dec/19

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

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

a_{n} = \frac{2}{T} \int_{[T]} ∣ x ∣ c o s (n x) = \frac{1}{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)

=_{b y p a r t s} {[\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} {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} {{(- 1)}^{n} - 1) \Rightarrow

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

a_{0} = \frac{2}{π} \times \frac{π^{2}}{2} = π \Rightarrow∣ 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{2}{π} (- 2) \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} c o s (2 p + 1) x

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

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

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

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \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}{3} \times \frac{π^{2}}{8} = \frac{π^{2}}{6} .