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Let-f-x-x-1-x-2-and-r-0-f-x-r-be-a-convergent-series-Find-the-value-of-x-such-that-n-0-r-0-x-1-x-2-r-n-4-




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

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