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Question Number 115910 by Eric002 last updated on 29/Sep/20

$$letxbeaposativerealnumber$$$$provethat$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n-1)!}{(x+1)....(x+n)}=\frac{1}{x}$$

Commented by Dwaipayan Shikari last updated on 29/Sep/20

$$$$$$\frac{1}{(\mathrm{x}+1)}+\frac{1!}{(\mathrm{x}+1)(\mathrm{x}+2)}+\frac{2!}{(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)}+....$$$$\mathrm{take}\mathrm{x}=1$$$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}(\mathrm{n}+1)}=\frac{1}{1}$$$$\mathrm{Which}\mathrm{is}\mathrm{true}$$$$$$

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