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Question Number 37859 by kunal1234523 last updated on 18/Jun/18

$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{\underset{i=0}{\prod ^{k-1}}(n-i)}{k!}$$$$\mathrm{is}\mathrm{this}\mathrm{really}\mathrm{mean}\mathrm{something}$$

Commented by math khazana by abdo last updated on 18/Jun/18

$$\prod _{i=0}^{k-1}(n-i)=n(n-1)(n-2)...(n-k+1)$$$$=\frac{n(n-1)....(n-k+1)(n-k)(n-k-1....2.1}{(n-k)!}$$$$=\frac{n!}{(n-k)!}\Rightarrow \frac{\prod _{i=0}^{k-1}(n-i)}{k!}=C_n^k\Rightarrow$$$$\sum _{k=1}^{\mathrm{\infty }}(...)=\sum _{k=1}^{\mathrm{\infty }}{C}_n^{k}andthisisanonsens$$$$becausek?mustbe⩽nbydefinitionof$$$$combination...$$

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