How to find the limit $\lim_{n\to \infty}\frac{n}{2^n-1}$? I've been trying to find and prove the 
$$\lim_{n\rightarrow\infty}\frac{n}{2^n-1}$$
but I haven't even figure out the limit, could anyone help?

Notice, 
$$ \frac{n}{2^n - 1} = \frac{\frac{n}{2^n}}{1 - \frac{1}{2^n}}$$
Obviosuly, $( \frac{1}{2} )^n \to 0 $.
All you have to do now is to show $ \frac{ n }{2^n} \to 0 $. 
Hint: For $n > 4 $ we have 
$$ \frac{n}{2^n} < \frac{n}{n^2} = \frac{1}{n}$$
