<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] If you multiply the top by $2$, you get $2x = 2+x$, whence $x = 2$
Likewise, the bottom multiplied by $3$ gives $3y = 3+y$, gives $y = \frac{3}{2}$.
So you then find $\left(\frac{2}{1.5}\right) = \frac{4}{3}$.
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