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Question Number 122036 by physicstutes last updated on 13/Nov/20

$$\mathrm{Let}f\left(x\right)=\left(\frac{x+1}{x+2}\right)\mathrm{and}\underset{r=0}{\sum ^{\mathrm{\infty }}}{\left[f\left(x\right)\right]}^r\mathrm{be}\mathrm{a}\mathrm{convergent}\mathrm{series}$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{such}\mathrm{that}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left[\underset{r=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{x+1}{x+2}\right)}^r\right]}^n=4$$

Answered by Dwaipayan Shikari last updated on 13/Nov/20

$$f\left(x\right)=\frac{x+1}{x+2}$$$$\underset{r=0}{\sum ^{\mathrm{\infty }}}{\left[f\left(x\right)\right]}^r=1+\frac{x+1}{x+2}+...=\frac{1}{1-\frac{x+1}{x+2}}=x+2$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{(x+2)}^n=\frac{1}{1-(x+2)}=\frac{1}{-(x+1)}$$$$\frac{1}{-(x+1)}=4\Rightarrow -(x+1)=\frac{1}{4}\Rightarrow x=-\frac{5}{4}$$$$$$

Commented by physicstutes last updated on 13/Nov/20

$$\mathrm{Fantastic}\mathrm{result}.\mathrm{What}\mathrm{i}\mathrm{execpted}.\mathrm{Thanks}\mathrm{alot}.$$

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