Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone The question was:

Guess the limit of the series, and prove your guess from the definition of limits (without any of the theorems thought in class).

I guessed that the limit is $0$, and then looking at the definition of a limit I have:
$\forall\epsilon>0\:\exists n_{0}>0\:\forall n>n_{0},\left|a_{n}-0\right|<\epsilon$
This means that I need to find $n_0$ such that for all $n>n_0$, $\frac{n_0}{2^{n_0}-1}<\epsilon$. But since I already prove that the sequence is decreasing, it's enough to find such $n_0$.
But... that's where I'm stuck...
 A: $$2^n = e^{n\log2 } = 1 + n \log 2 + \dfrac{n^2 \log^2(2)}2 + \cdots \geq 1 + \dfrac{n^2 \log^2(2)}2$$
Hence, we get that $$2^n - 1 \geq \dfrac{n^2 \log^2(2)}2$$
Hence, $$\dfrac{n}{2^n-1} \leq \dfrac{2n}{n^2 (\log 2)^2} = \dfrac{2}{n (\log 2)^2}$$
Now choose $N = \left \lceil \dfrac{2}{\epsilon (\log 2)^2} \right \rceil$. Then we have that for $n \geq N$
$$\dfrac{n}{2^n-1} \leq \dfrac{2}{n (\log 2)^2} < \epsilon$$
EDIT
If you are not comfortable with $\log$, then here is another way out. Prove that
$$2^n \geq 1 + \dfrac{n^2}3$$ using induction. Hence, we now have that $$\dfrac{n}{2^n-1} \leq \dfrac3n$$
Now given $\epsilon > 0$ choose $N = \left \lceil \dfrac3 \epsilon\right \rceil$. Hence, for all $n > N$, we have that $$\dfrac{n}{2^n-1} \leq \dfrac3n < \epsilon$$
A: Hint:
Use induction to prove that $2^n-1>n^2 \ \forall n \geq 5 $.
Conclude that $\dfrac{n}{2^n-1}<\dfrac{1}{n} \ \forall n \geq5$.
Now it should be easy. For $\epsilon>0$ if $n_0=\max{\{5,\lceil\frac{1}{\epsilon}\rceil+1\}}$ then $\dfrac{n}{2^n-1}<\epsilon \ \ \forall n \geq n_0$.
A: It all comes down to showing that $2^n-1$ is "big" compared to $n$. This is the intuition that led you to decide that the limit is $0$. But we need to fill in some details. 
We can do it by using the Binomial Theorem. Imagine expanding $(1+1)^n$, where $n\ge 2$. The first three terms in the expansion are $1$, $n$, and $\dfrac{n(n-1)}{2}$.
It follows that if $n\ge 2$,
$$2^n \gt \frac{n(n-1)}{2}.$$
Thus for $n\ge 2$, the $n$-th term of your sequence is $\le \dfrac{2}{n-1}$. Now it should not be difficult, given $\epsilon$, to find an $n_0$ that works. 
A: Hints: 


*

*$n<1.5^{n}$

*$2^{n}-1>2^{n-1}$

*Use $\log_{\text{something}}(\phantom{x})$

