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Question Number 167663 by mathlove last updated on 22/Mar/22

lim_(x→2) [log((1/x)+(1/(2x))+(1/(4x)).......)]=?

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left[{log}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}.......\right)\right]=? \\ $$

Commented by mathlove last updated on 22/Mar/22

????

$$???? \\ $$

Commented by alephzero last updated on 22/Mar/22

did You mean (1/x)+(1/(2x))+(1/(4x))+...=(1/x)+(1/(2x))+...+(1/(2nx))+... or (1/x)+(1/(2x))+(1/(4x))+...=(1/x)+(1/(2x))+...+(1/(2^n x))+...?

$${did}\:{You}\:{mean}\: \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}+...=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+...+\frac{\mathrm{1}}{\mathrm{2}{nx}}+... \\ $$$${or} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}+...=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} {x}}+...? \\ $$

Answered by alephzero last updated on 22/Mar/22

(1/(2^0 x))+(1/(2^1 x))+(1/(2^2 x))+...=Σ_(n=0) ^∞ (1/(2^n x)) S_n = ((a_1 (1−r^n ))/(1−r)) a_1 = (1/x) ∧ r = (1/2) ⇒ S = S_∞ = ((((1/x)))/(((1/2)))) = (2/x) ⇒ lim_(x→2) ln((1/x)+(1/(2x))+...) = = lim_(x→2) ln((2/x)) = 0

$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} {x}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} {x}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} {x}}+...=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} {x}} \\ $$$${S}_{{n}} \:=\:\frac{{a}_{\mathrm{1}} \left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$${a}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{{x}}\:\wedge\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{S}\:=\:{S}_{\infty} \:=\:\frac{\left(\frac{\mathrm{1}}{{x}}\right)}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}}{{x}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}ln}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+...\right)\:= \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}ln}\left(\frac{\mathrm{2}}{{x}}\right)\:=\:\mathrm{0} \\ $$

Commented by mathlove last updated on 22/Mar/22

thanks sir

$${thanks}\:{sir} \\ $$

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