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Question Number 175860 by Shrinava last updated on 08/Sep/22

Answered by mr W last updated on 08/Sep/22

we know: (1/1^2 )+(1/2^2 )+(1/3^2 )+(1/4^2 )+...=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 )+...)+((1/2^2 )+(1/4^2 )+(1/6^2 )+...)=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 )+...)+(1/4)((1/1^2 )+(1/2^2 )+(1/3^2 )+...)=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 )+...)+(1/4)×(π^2 /6)=(π^2 /6) (1/1^2 )+(1/3^2 )+(1/5^2 )+...=(π^2 /8) Σ_(n=1) ^∞ (1/((2n−1)^2 ))=(π^2 /8) Ω=Σ_(n=1) ^∞ ((4n^2 +1)/((4n^2 −1)^2 )) =(1/2)Σ_(n=1) ^∞ [(1/((2n−1)^2 ))+(1/((2n+1)^2 ))] =(1/2)Σ_(n=1) ^∞ [(1/((2n−1)^2 ))+(1/((2n−1)^2 ))]−(1/2) =Σ_(n=1) ^∞ (1/((2n−1)^2 ))−(1/2) =(π^2 /8)−(1/2) ✓

weknow:112+122+132+142+...=π26(112+132+152+...)+(122+142+162+...)=π26(112+132+152+...)+14(112+122+132+...)=π26(112+132+152+...)+14×π26=π26112+132+152+...=π28∑∞n=11(2n−1)2=π28Ω=∑∞n=14n2+1(4n2−1)2=12∑∞n=1[1(2n−1)2+1(2n+1)2]=12∑∞n=1[1(2n−1)2+1(2n−1)2]−12=∑∞n=11(2n−1)2−12=π28−12✓

Commented by Tawa11 last updated on 08/Sep/22

Great sir

Greatsir

Answered by mnjuly1970 last updated on 08/Sep/22

Ω= Σ_(n=1) ^∞ (( (2n+1)^( 2) −4n)/((2n+1)^( 2) (2n−1)^( 2) )) = Σ_(n=1) ^∞ (1/((2n−1)^( 2) )) −{(1/2)Σ_(n=1) ^∞ (1/((2n−1)^2 )) −(1/((2n+1)^( 2) ))} = (1/2)Σ_(n=1) ^∞ (( 1)/((2n−1)^( 2) )) +(1/2) Σ_(n=1) ^∞ (1/((2n−1)^( 2) )) −(1/2) =Σ_(n=1) ^∞ (1/((2n−1)^( 2) )) −(1/2)= (π^( 2) /8) −(1/2)

Ω=∑∞n=1(2n+1)2−4n(2n+1)2(2n−1)2=∑∞n=11(2n−1)2−{12∑∞n=11(2n−1)2−1(2n+1)2}=12∑∞n=11(2n−1)2+12∑∞n=11(2n−1)2−12=∑∞n=11(2n−1)2−12=π28−12

Commented by Tawa11 last updated on 15/Sep/22

Great sir

Greatsir

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