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Question Number 138437 by EnterUsername last updated on 13/Apr/21

Σ_(n=0) ^∞ (1/((2n)!!))=(√e)

$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\sqrt{\mathrm{e}} \\ $$

Answered by Ar Brandon last updated on 13/Apr/21

(2n)!!=2n(2n−2)(2n−4)...2 =2^n n(n−1)(n−2)...1=2^n n! Σ_(n=0) ^∞ (1/((2n)!!))=Σ_(n=0) ^∞ (1/(2^n n!))=Σ_(n=0) ^∞ (((1/2)^n )/(n!))=e^(1/2) =(√e)

$$\:\:\:\:\:\left(\mathrm{2n}\right)!!=\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(\mathrm{2n}−\mathrm{4}\right)...\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{n}} \mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)...\mathrm{1}=\mathrm{2}^{\mathrm{n}} \mathrm{n}! \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} \mathrm{n}!}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{e}} \\ $$

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