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Question Number 203180 by Calculusboy last updated on 11/Jan/24

Answered by Rana_Ranino last updated on 11/Jan/24

$$\mathrm{using}{\arcsin }^2\left(\mathrm{z}\right)=\frac{1}{2}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{4^{\mathrm{n}}{\mathrm{z}}^{2\mathrm{n}}}{{\mathrm{n}}^2\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)}\mathrm{take}\mathrm{z}=\frac{1}{2}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{4^{\mathrm{n}}{\mathrm{z}}^{2\mathrm{n}}}{{\mathrm{n}}^2\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)}={2\arcsin }^2\left(\frac{1}{2}\right)=2{\left(\frac{\pi }{6}\right)}^2=\frac{{\pi }^2}{18}$$$$$$

Commented by Calculusboy last updated on 11/Jan/24

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{m}\mathit{a}$$

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