<s>[INST] A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:

$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$

What I've done is to simplify the upper part to: $$\frac{2^{n+1}-1}{2^n}$$
Any hints or solutions will be greatly appreciated.
 [/INST] Here are the steps
$$ \lim\limits_{n\to\infty} \frac{\sum\limits_{k=0}^n \frac{1}{2^k}}{\sum\limits_{k=0}^n \frac{1}{3^k}} = \lim\limits_{n\to\infty} \frac{2-\frac1{2^n}}{\frac12\left(3-\frac1{3^n}\right)}$$
$$= \frac{2-0}{\frac12\left(3-0\right)} = \frac2{\frac32} = \frac43$$
Study geometric series.
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