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Question Number 121875 by rs4089 last updated on 12/Nov/20

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\frac{{\pi }^4}{90}$$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\frac{sin\pi x}{\pi x}=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{x^2}{n^2})$$$$\frac{sinu}{u}=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{u^2}{{\pi }^2{n}^2})$$$$1-\frac{u^2}{6}+\frac{u^4}{120}-...=(1-\frac{u^2}{{\pi }^2})(1-\frac{u^2}{4{\pi }^2})(1-\frac{u^2}{9{\pi }^2})(1-\frac{u^2}{16{\pi }^2})...$$$$1-\frac{u^2}{6}+\frac{u^4}{120}-...=1-(\frac{u^2}{{\pi }^2}+\frac{u^2}{4{\pi }^2}+...)+\frac{3}{4}(\frac{u^4}{{\pi }^4}+\frac{u^4}{16{\pi }^4}+...)+....$$$$\frac{u^2}{6}=u^2(\frac{1}{{\pi }^2}+\frac{1}{4{\pi }^2}+...)$$$$1+\frac{1}{4}+\frac{1}{9}+..=\frac{{\pi }^2}{6}$$$$\frac{u^4}{120}=\frac{3}{4}(\frac{u^4}{{\pi }^4}+\frac{u^4}{16{\pi }^4}+.....)$$$$1+\frac{1}{16}+\frac{1}{81}+...=\frac{{\pi }^4}{90}$$

Commented by mnjuly1970 last updated on 12/Nov/20

$$niceverynice$$$$withoutusingthefourierseries$$$$thankyou...$$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

�� with pleasure sir!

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