Question Number 44301 by naka3546 last updated on 26/Sep/18
$${Prove}\:\:{that} \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\:+\:…\:\:=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
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) .](https://www.tinkutara.com/question/Q44310.png)
$${let}\:{consider}\:{f}\left({x}\right)=\mid{x}\mid\:\left(\mathrm{2}\pi{periodic}\:{even}\right) \\ $$$${let}\:{developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right) \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \mid{x}\mid{cos}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx} \\ $$$$=\left[\frac{{x}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\:\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left\{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{\pi}\:\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}{a}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\pi} \:{x}\:{dx}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\pi} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}} \\ $$$$\mid{x}\mid\:=\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{2}} }\:{cos}\left({nx}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{4}}{\pi}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{n}+\mathrm{1}\right){x}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}=\mathrm{0}\:\Rightarrow\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{4}}{\pi}\:\sum_{{n}=\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{4}}{\pi}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${but}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$