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?
 A: $$\lim_{n\to\infty}\frac{n}{2^n-1}=\lim_{n\to\infty}\frac{1}{2^n\ln2}=\dfrac{1}{\ln{2}}\lim_{n\to\infty}2^{-n}=\dfrac{1}{\ln{2}}\cdot0=0$$
A: 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 
$$ \frac{n}{2^n} < \frac{n}{n^2} = \frac{1}{n}$$
A: From a certain $n$, $2^n-1>n^2$, therefore $$0\leq \frac{n}{2^n-1}\leq \frac{n}{n^2}=\frac{1}{n}\to 0$$
A: Let
$$u_n=2^n-n^2-1$$
And
$$v_n=u_{n+1}-u_n=2^{n+1}-2^n-(n+1)^2+n^2=2^n-2n-1$$
Then$$v_{n+1}-v_n=2^n-2$$
Hence $v_{n+1}-v_n\geq2$ as long as $n\geq2$, and $v_3=1$, hence $v_n\geq 2(n-3)+1>0$ for $n\geq3$.
Thus the sequence $u_n$ is inceasing for $n\geq3$, and $u_5=32-25-1=6$, thus $u_n>0$ for $n\geq5$.
That is, $2^n-1> n^2$ for $n\geq5$.
Then, for $n\geq5$,
$$\frac{n}{2^n-1}<\frac{n}{n^2}=\frac1n\underset{n\to\infty}\longrightarrow0$$
A: For $a>1$, we have $a^n = e^{n \log a} = \displaystyle \sum_{k=0}^{\infty} \dfrac{(n \log a)^k}{k!} \geq 1+\dfrac{n^2 \log^2(a)}{2!}$. Hence, we have
$$0 \leq \dfrac{n}{a^n-1} \leq \dfrac2{n \log^2(a)} \to 0$$
A: You can use Stolz theorem:
$ \lim \frac{a_n}{b_n}  = \lim\frac{a_{n+1}- a_n}{b_{n+1}-b_n} $
provided that $ b_n \rightarrow \infty $ monotonously
