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Question-135786




Question Number 135786 by JulioCesar last updated on 16/Mar/21
Commented by Ar Brandon last updated on 16/Mar/21
You mean   H=lim_(n→∞) ((1+(1/2)+(1/4)+∙∙∙+(1/2^n ))/(1+(1/3)+(1/9)+∙∙∙+(1/3^n ))) ???
$$\mathrm{You}\:\mathrm{mean}\: \\ $$$$\mathrm{H}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}\:??? \\ $$
Answered by Ar Brandon last updated on 16/Mar/21
H=((Σ_(n=0) ^∞ (1/2^n ))/(Σ_(n=0) ^∞ (1/3^n )))=((1−(1/3))/(1−(1/2)))=(2/3)∙(2/1)=(4/3)
$$\mathrm{H}=\frac{\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }}{\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$

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