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Question Number 37712 by Rio Mike last updated on 16/Jun/18

$$Evaluate\underset{r=0}{\sum ^{\mathrm{\infty }}}{2}^{r-1}$$

Commented by prakash jain last updated on 17/Jun/18

$$-1/2\mathrm{is}\mathrm{also}\mathrm{a}\mathrm{valid}\mathrm{answer}\mathrm{using}$$$$\mathrm{analytical}\mathrm{continuity}$$

Answered by Joel579 last updated on 17/Jun/18

$$S_n=2^{-1}+2^0+2^1+2^2+2^3+2^4+...+2^{n-1}$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n=\mathrm{\infty }$$

Commented by Rio Mike last updated on 17/Jun/18

$$$$$$\underset{r=0}{\sum ^{\mathrm{\infty }}}{2}^{r-1}$$$$2^{0-1}+2^{1-1}+2^{2-1}+2^{3-1}+...2^{r-n}$$$$GP=\frac{1}{2}+1+2+4+...$$$$a=1/2andr=2$$$$S_{\mathrm{\infty }}=\frac{a}{1-r}$$$$=\frac{\frac{1}{2}}{1-2}$$$$=-1/2$$$$={}_{}\underset{n\to 0}{\lim }{S}_n=\mathrm{\infty }......$$

Commented by math1967 last updated on 17/Jun/18

$$For\frac{a}{1-r}r<1thenwhyr=2applicable?$$

Commented by Joel579 last updated on 17/Jun/18

$$S_{\mathrm{\infty }}=\frac{a}{1-r}\mathrm{only}\mathrm{for}\mid r\mid <1$$

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