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Question Number 135944 by mnjuly1970 last updated on 17/Mar/21

.....nice calculus... prove that: Σ_(n=0) ^∞ ((((√2) −1)^n )/((2n+1)^2 ))=(π^2 /8)−(1/2)ln^2 (1+(√2) )... ................✓

.....nicecalculus...provethat:∑∞n=0(2−1)n(2n+1)2=π28−12ln2(1+2)...................✓

Answered by mindispower last updated on 19/Mar/21

hello sir sur for this result ? we can Σ_(n≥0) ((((√2)−1)^n )/((2n+1)^2 ))=(1/2)Σ_(n≥0) 2((((√((√2)−1)))^(2n) )/((2n+1)^2 )) =(1/(2.(√((√2)−1))))Σ_(n≥0) ((((√((√2)−1)))^(2n+1) )/((2n+1)^2 )) =(1/(2(√((√2)−1))))(Σ_(n≥1) ((((√((√2)−1)))^n )/n^2 )−Σ_(n≥1) (((−(√((√2)−1)))^n )/n^2 )) =(1/(2(√((√2)−1))))(Li_2 ((√((√2)−1)))−Li_2 (−(√((√2)−1)))

hellosirsurforthisresult?wecan∑n⩾0(2−1)n(2n+1)2=12∑n⩾02(2−1)2n(2n+1)2=12.2−1∑n⩾0(2−1)2n+1(2n+1)2=122−1(∑n⩾1(2−1)nn2−∑n⩾1(−2−1)nn2)=122−1(Li2(2−1)−Li2(−2−1)

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