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

let x be a posative real number  prove that  Σ_(n=1) ^∞ (((n−1)!)/((x+1)....(x+n)))=(1/x)

$${let}\:{x}\:{be}\:{a}\:{posative}\:{real}\:{number} \\ $$$${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)....\left({x}+{n}\right)}=\frac{\mathrm{1}}{{x}} \\ $$

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

  (1/((x+1)))+((1!)/((x+1)(x+2)))+((2!)/((x+1)(x+2)(x+3)))+....  take x=1  (1/2)+(1/6)+(1/(12))+...=Σ_(n=1) ^∞ (1/(n(n+1)))=(1/1)  Which is true

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

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