Prove that 1 + (1/2^2 ) + (1/3^2 ) + (1/4^2 ) + ... = (π^2 /6)


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

$P r o v e t h a t$ $1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . = \frac{π^{2}}{6}$
Commented by abdo.msup.com last updated on 26/Sep/18
$let consider f(x)=∣x∣ (2πperiodic even) let developp f at fourier serie f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T) ∫_([T]) f(x)cos(nx)dx =(2/(2π)) ∫_(−π) ^π ∣x∣cos(nx)dx =(2/π) ∫_0 ^π x cos(nx)dx ⇒(π/2)a_n =∫_0 ^π x cos(nx)dx =[(x/n)sin(nx)]_0 ^π −∫_0 ^π (1/n)sin(nx)dx =−(1/n) ∫_0 ^π sin(nx)dx =−(1/n) [−(1/n)cos(nx)]_0 ^π =(1/n^2 ){(−1)^n −1} ⇒(π/2)a_n =(((−1)^n −1)/n^2 ) a_n =(2/π) (((−1)^n −1)/n^2 ) ⇒ (π/2)a_0 =∫_0 ^π x dx=[(x^2 /2)]_0 ^π =(π^2 /2) ⇒(a_0 /2) =(π/2) ∣x∣ =(π/2) +(2/π)Σ_(n=1) ^∞ (((−1)^n −1)/n^2 ) cos(nx) =(π/2) −(4/π) Σ_(n=0) ^∞ ((cos(2n+1)x)/((2n+1)^2 )) x=0 ⇒(π/2) −(4/π) Σ_(n=0) (1/((2n+1)^2 )) ⇒ (4/π) Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π/2) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 ))=(π^2 /8) but Σ_(n=1) ^∞ (1/n^2 ) =Σ_(p=1) ^∞ (1/((2p)^2 )) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +(π^2 /8) ⇒ (3/4)Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /8) ⇒Σ_(n=1) ^∞ (1/n^2 ) =((4π^2 )/(24)) =(π^2 /6) .$
$l e t c o n s i d e r f (x) =∣ x ∣ (2 π p e r i o d i c e v e n)$ $l e t d e v e l o p p f a t f o u r i e r s e r i e$ $f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x)$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} ∣ x ∣ c o s (n x) d x = \frac{2}{π} \int_{0}^{π} x c o s (n x) d x$ $\Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x c o s (n x) d x$ $= {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x$ $= - \frac{1}{n} \int_{0}^{π} s i n (n x) d x$ $= - \frac{1}{n} {[- \frac{1}{n} c o s (n x)]}_{0}^{π}$ $= \frac{1}{n^{2}} {{(- 1)}^{n} - 1} \Rightarrow \frac{π}{2} a_{n} = \frac{{(- 1)}^{n} - 1}{n^{2}}$ $a_{n} = \frac{2}{π} \frac{{(- 1)}^{n} - 1}{n^{2}} \Rightarrow$ $\frac{π}{2} a_{0} = \int_{0}^{π} x d x = {[\frac{x^{2}}{2}]}_{0}^{π} = \frac{π^{2}}{2} \Rightarrow \frac{a_{0}}{2} = \frac{π}{2}$ $∣ x ∣ = \frac{π}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} - 1}{n^{2}} c o s (n x)$ $= \frac{π}{2} - \frac{4}{π} \sum_{n = 0}^{\infty} \frac{c o s (2 n + 1) x}{{(2 n + 1)}^{2}}$ $x = 0 \Rightarrow \frac{π}{2} - \frac{4}{π} \sum_{n = 0} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\frac{4}{π} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π}{2} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}$ $b u t \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{p = 1}^{\infty} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}}$ $= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \frac{π^{2}}{8} \Rightarrow$ $\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4 π^{2}}{24}$ $= \frac{π^{2}}{6} .$
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