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