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n-0-5-2-2n-1-2n-1-2-pi-2-24-1-12-log-2-2-5-




Question Number 135638 by Dwaipayan Shikari last updated on 14/Mar/21
Σ_(n=0) ^∞ ((((√5)−2)^(2n+1) )/((2n+1)^2 ))=(π^2 /(24))−(1/(12))log^2 (2+(√5))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{12}}{log}^{\mathrm{2}} \left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$

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