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Question Number 176054 by BaliramKumar last updated on 11/Sep/22
(1/(2!)) + (2/(3!)) + (3/(4!)) + (4/(5!)) + ............. ∞ = ?
$$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:\frac{\mathrm{4}}{\mathrm{5}!}\:+\:………….\:\infty\:=\:? \\ $$
Answered by Ar Brandon last updated on 11/Sep/22
S=Σ_(n=2) ^∞ ((n−1)/(n!))=Σ_(n=2) ^∞ (1/((n−1)!))−Σ_(n=2) ^∞ (1/(n!))    =Σ_(n=1) ^∞ (1/(n!))−Σ_(n=2) ^∞ (1/(n!))=Σ_(n=1) ^∞ (1/(n!))−(Σ_(n=1) ^∞ (1/(n!))−1)=1
$${S}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}−\mathrm{1}}{{n}!}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!} \\ $$$$\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right)=\mathrm{1} \\ $$

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