A limit with geometrical progression I got stuck trying to solve this limit. 
$$ \lim_{n\to\infty} \frac{1 + \frac{1}{2} + \frac{1}{4} + ....+ \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}} $$
The first thing I did was calculating the sums of the geometrical progressions found in both numerator and denumerator. And after that I got
$$ \lim_{n\to\infty} \frac{\frac{(\frac{1}{2})^n - 1}{\frac{1}{2} - 1}}{\frac{(\frac{1}{3})^n - 1}{\frac{1}{3} - 1}} $$
And here I am stuck because I don't know how to continue. Any help?

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
$$
\lim_{n\to\infty}\frac{\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)}{\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)}=\frac{2}{\frac{3}{2}}=\frac{4}{3}.
$$
