prove that Σ_(r=1) ^∞ (1/r^2 ) = (π^2 /6)


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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
$let f(x)=∣x∣ ,2π periodic even f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) ∣x∣ cos(nx) =(1/π)∫_(−π) ^π ∣x∣ cos(nx)dx =(2/π) ∫_0 ^π x cos(nx)dx ⇒(π/2)a_n =∫_0 ^π x cos(nx) =_(by parts) [(x/n)sin(nx)]_0 ^π −∫_0 ^π (1/n)sin(nx)dx =−(1/n)[−(1/n)cos(nx)]_0 ^π =(1/n^2 ){(−1)^n −1) ⇒ a_n =(2/(πn^2 )){ (−1)^n −1} a_0 =(2/π)×(π^2 /2)=π ⇒∣x∣ =(π/2) +(2/π)Σ_(n=1) ^∞ (((−1)^n −1)/n^2 )cos(nx) =(π/2) +(2/π)(−2) Σ_(p=0) ^∞ (1/((2p+1)^2 ))cos(2p+1)x =(π/2)−(4/π) Σ_(p=0) ^∞ ((cos(2p+1)x)/((2p+1)^2 )) x=0 ⇒(π/2)−(4/π) Σ_(p=0) ^∞ (1/((2p+1)^2 )) =0 ⇒(4/π)Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(π/2) ⇒ Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(π^2 /8) Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ (3/4) Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /8) ⇒ Σ_(n=1) ^∞ (1/n^2 ) =(4/3)×(π^2 /8) =(π^2 /6) .$
$l e t f (x) =∣ x ∣, 2 π p e r i o d i c e v e n$ $f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x)$ $a_{n} = \frac{2}{T} \int_{[T]} ∣ x ∣ c o s (n x) = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ c o s (n x) d x$ $= \frac{2}{π} \int_{0}^{π} x c o s (n x) d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x c o s (n x)$ $=_{b y p a r t s} {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x$ $= - \frac{1}{n} {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} {{(- 1)}^{n} - 1) \Rightarrow$ $a_{n} = \frac{2}{π n^{2}} {{(- 1)}^{n} - 1}$ $a_{0} = \frac{2}{π} \times \frac{π^{2}}{2} = π \Rightarrow∣ x ∣ = \frac{π}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} - 1}{n^{2}} c o s (n x)$ $= \frac{π}{2} + \frac{2}{π} (- 2) \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} c o s (2 p + 1) x$ $= \frac{π}{2} - \frac{4}{π} \sum_{p = 0}^{\infty} \frac{c o s (2 p + 1) x}{{(2 p + 1)}^{2}}$ $x = 0 \Rightarrow \frac{π}{2} - \frac{4}{π} \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} = 0 \Rightarrow \frac{4}{π} \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} = \frac{π}{2} \Rightarrow$ $\sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} = \frac{π^{2}}{8}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4}{3} \times \frac{π^{2}}{8} = \frac{π^{2}}{6} .$
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