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

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

Provethat1+122+132+142+...=π26

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) .

letconsiderf(x)=∣x(2πperiodiceven)letdeveloppfatfourierserief(x)=a02+n=1ancos(nx)an=2T[T]f(x)cos(nx)dx=22πππxcos(nx)dx=2π0πxcos(nx)dxπ2an=0πxcos(nx)dx=[xnsin(nx)]0π0π1nsin(nx)dx=1n0πsin(nx)dx=1n[1ncos(nx)]0π=1n2{(1)n1}π2an=(1)n1n2an=2π(1)n1n2π2a0=0πxdx=[x22]0π=π22a02=π2x=π2+2πn=1(1)n1n2cos(nx)=π24πn=0cos(2n+1)x(2n+1)2x=0π24πn=01(2n+1)24πn=01(2n+1)2=π2n=01(2n+1)2=π28butn=11n2=p=11(2p)2+p=01(2p+1)2=14n=11n2+π2834n=11n2=π28n=11n2=4π224=π26.

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