A limit with geometrical progression

We see, that $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+ \frac{1}{2^{n}}\right)$ and $\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)$ are the geometric series.

Therefore, $$ \lim_{n\to\infty}\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\frac{1}{1-\frac{1}{2}}={2}, $$ $$ \lim_{n\to\infty}\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n=\frac{1}{1-\frac{1}{3}}=\frac{3}{2} $$ We will have $$ \lim_{n\to\infty}\frac{\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)}{\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)}=\frac{2}{\frac{3}{2}}=\frac{4}{3}. $$

Begin from the last step that you have reached. Write it as two separate fractions so it is "nicer": $$= \lim_{n\to\infty}\frac{(\frac{1}{2})^n - 1}{\frac{1}{2} - 1}\times\frac{\frac{1}{3} - 1}{(\frac{1}{3})^n - 1}$$

As $n\rightarrow\infty, (\frac{1}{2})^n\rightarrow 0, (\frac{1}{3})^n\rightarrow 0$

$$=\lim_{n\to\infty}\frac{- 1}{\frac{-1}{2}}\times\frac{-2/3}{-1}=\lim_{n\to\infty}\frac{4}{3}=\frac{4}{3}$$