Question Number 152496 by mathdanisur last updated on 28/Aug/21
Answered by qaz last updated on 29/Aug/21
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{4n}} }{\left(\mathrm{4n}\right)!}=\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{8}!}+\frac{\mathrm{x}^{\mathrm{12}} }{\mathrm{12}!}+… \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}−…\right)+\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+…\right)\right]−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosh}\:\mathrm{x}−\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{4n}\right)!}\left(\frac{\pi}{\mathrm{6}}\right)^{\mathrm{4n}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosh}\:\frac{\pi}{\mathrm{6}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}^{\pi/\mathrm{6}} +\mathrm{e}^{−\pi/\mathrm{6}} \right)−\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 29/Aug/21
$${thankyou}\:{Ser} \\ $$