Question Number 122036 by physicstutes last updated on 13/Nov/20
$$\mathrm{Let}\:\:{f}\left({x}\right)\:=\:\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)\:\mathrm{and}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\left[{f}\left({x}\right)\right]^{{r}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{convergent}\:\mathrm{series} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)^{{r}} \right]^{{n}} \:=\:\mathrm{4}\: \\ $$
Answered by Dwaipayan Shikari last updated on 13/Nov/20
$${f}\left({x}\right)=\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\left[{f}\left({x}\right)\right]^{{r}} =\mathrm{1}+\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}+…=\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}={x}+\mathrm{2} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({x}+\mathrm{2}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{−\left({x}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{−\left({x}+\mathrm{1}\right)}=\mathrm{4}\Rightarrow−\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{x}=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by physicstutes last updated on 13/Nov/20
$$\mathrm{Fantastic}\:\mathrm{result}.\:\mathrm{What}\:\mathrm{i}\:\mathrm{execpted}.\:\mathrm{Thanks}\:\mathrm{alot}. \\ $$