$\underset{n\to \infty }{Lim}\frac{1+\frac{1}{2}+\frac{1}{4}+\mathrm{....}+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{9}+\mathrm{....}+\frac{1}{3^n}}=$

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

$\underset{n\to \infty }{\lim }\frac{\frac{1\left(1-{\left(\frac{1}{2}\right)}^n\right)}{1-\frac{1}{2}}}{\frac{1-{\left(\frac{1}{3}\right)}^n}{1-\frac{1}{3}}}$

$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim} \frac{\frac{2^{n-1}}{2^n}}{\frac{1}{2}}\times \frac{\frac{2}{3}}{\frac{3^{n-1}}{3^n}}\mid \\ \underset{n\to \mathrm{\infty }}{lim} \frac{2\left(2^{n-1}\right)}{2^n}\times \frac{2(3{)}^n}{3\left(3^{n-1}\right)}\\ =\frac{4}{3}\underset{n\to \mathrm{\infty }}{lim} \left(\frac{3}{2}\right)\left(\frac{2^{n-1}}{3^{n-1}}\right)\\ =\frac{4}{3}\underset{n\to \mathrm{\infty }}{lim} \frac{3^n}{2^n}\frac{2^n\left(1-\frac{1}{2^n}\right)}{3^n\left(1-\frac{1}{3^n}\right)}\end{array}$

$\begin{array}{l}=\frac{4}{3}\underset{x\to \mathrm{\infty }}{lim} \frac{1-\frac{1}{2^n}}{1-\frac{1}{3^n}}\\ =\frac{4}{3}\end{array}$