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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) )...      ................✓
$$\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:{calculus}… \\ $$$$\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)… \\ $$$$\:\:\:\:…………….\checkmark \\ $$
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)))
$${hello}\:{sir}\:{sur}\:\:{for}\:{this}\:{result}\:? \\ $$$${we}\:{can} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\mathrm{2}\frac{\left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}.\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{{n}} }{{n}^{\mathrm{2}} }−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{{n}} }{{n}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}\left({Li}_{\mathrm{2}} \left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)−{Li}_{\mathrm{2}} \left(−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)\right. \\ $$

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