Question Number 138437 by EnterUsername last updated on 13/Apr/21 | ||
$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\sqrt{\mathrm{e}} \\ $$ | ||
Answered by Ar Brandon last updated on 13/Apr/21 | ||
$$\:\:\:\:\:\left(\mathrm{2n}\right)!!=\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(\mathrm{2n}−\mathrm{4}\right)...\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{n}} \mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)...\mathrm{1}=\mathrm{2}^{\mathrm{n}} \mathrm{n}! \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} \mathrm{n}!}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{e}} \\ $$ | ||