Question

${lim}_{n\to \infty }\frac{1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{3^n}}$

Answer 1

${lim}_{n\to \infty }\frac{1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^n}}$

$={lim}_{n\to \infty }\frac{\frac{1(1-\frac{1}{2^n})}{1-\frac{1}{2}}}{\frac{1(1-\frac{1}{3^n})}{1-\frac{1}{3}}}$

$={lim}_{n\to \infty }\frac{4(1-\frac{1}{2^n})}{3(1-\frac{1}{3^n})}$

$=\frac{4}{3}$

Therefore,

${lim}_{n\to \infty }\frac{1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^n}}=\frac{4}{3}$

Hence the correct answer is $\frac{4}{3}$.