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Question Number 152496 by mathdanisur last updated on 28/Aug/21

Answered by qaz last updated on 29/Aug/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{4\mathrm{n}}}{\left(4\mathrm{n}\right)!}=\frac{{\mathrm{x}}^4}{4!}+\frac{{\mathrm{x}}^8}{8!}+\frac{{\mathrm{x}}^{12}}{12!}+...$$$$=\frac{1}{2}[(1-\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}-...)+(1+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}+...)]-1$$$$=\frac{1}{2}\cos \mathrm{x}+\frac{1}{2}\cosh \mathrm{x}-1$$$$\Rightarrow \underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(4\mathrm{n}\right)!}{\left(\frac{\pi }{6}\right)}^{4\mathrm{n}}=\frac{1}{2}\cos \frac{\pi }{6}+\frac{1}{2}\cosh \frac{\pi }{6}-1=\frac{1}{4}({\mathrm{e}}^{\pi /6}+{\mathrm{e}}^{-\pi /6})-\frac{4-\sqrt{3}}{4}$$

Commented by mathdanisur last updated on 29/Aug/21

$$thankyouSer$$

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