Solving exponential inequality, epsilion delta proof I am doing a recap of what I've learned this year, I am trying to prove $\lim_{n \to \infty}{\frac{n}{2^n}} = 0$, using the definiton of limit (no integrals, derivatives, squeeze theorem, series etc.)
It boils down to this inequality $$\ln{n} - n\ln{2} < ln{\epsilon}$$
My initial thought was to use substitution $t = \ln{n}$, but after some time I end up with my initial inequality. Binomial Theorem did not help either. I keep going in circles, no matter what I do. Can someone give me a hint (no full solutions please, i suck at solving these kind of inequalities and want to improve).
 A: No, you don't need to solve that equation. You only need to find an upper bound and let this upper bound less than $\epsilon$.
For example: $x<(\frac{3}{2})^x$ for all $x\in(0,\infty)$
So, you have $\frac{n}{2^n}\le \frac{(3/2)^n}{2^n}\le (\frac{3}{4})^n\le \epsilon$
Therefore, you can take $N=\lfloor{\frac{\ln\epsilon}{\ln(3/4)}}\rfloor+1$
Can you proceed from here?
A: Let $\epsilon>0$ be given.
$|\frac{n}{2^n}|<\frac{n}{n^2}=\frac{1}{n}$$\quad \forall n\ge  5$
Then set $N=\max(\lfloor\frac{1}{\epsilon}\rfloor+1,5)$
Now $\forall \epsilon >0$ ,$\exists N \in\Bbb{N} $ such that $\forall n\ge N \implies $ $|\frac{n}{2^n}|<\epsilon$
Hence $(\frac{n}{2^n})$ converges to $0$.
A: $\underline{\text{Lemma 1}}$
Let $a_n$ denote $\dfrac{2^n}{n} ~: ~n \in \Bbb{Z^+}.$ 
Then, the subsequence $a_4, a_5, a_6, \cdots $ 
is strictly increasing, strictly positive, and grows unbounded. 
Proof
$a_4 = 4.$ 
For $~\displaystyle n \geq 4,$ 
$\displaystyle a_{n+1} = a_n \times \left[\frac{2n}{n+1}\right] = a_n \times 2 \times \left[1 - \frac{1}{n+1}\right] \geq a_n \times (1.6).$ 
Therefore, for all $~k \in \Bbb{Z^+}, ~a_{4+k} \geq 4 \times (1.6)^k.$ 
Therefore, the subsequence $a_4, a_5, a_6, \cdots,$ is strictly positive, strictly increasing, and
grows unbounded.

For any $\epsilon > 0,$ Lemma 1 implies that there exists an $~N \in \Bbb{Z^+},~$ such that for all $~n \geq N ~: n \in \Bbb{Z^+}, ~~\dfrac{2^n}{n} > \dfrac{1}{\epsilon} \implies \dfrac{n}{2^n} < \epsilon.$
Thus, the requirement necessary to conclude that 
$\displaystyle \lim_{n\to \infty} \dfrac{n}{2^n} = 0$ 
has been met.
A: Hint: look at the row $c_n=\frac{n}{2^n}$ and in order to show that it converges to $0$, you want to show that for all $\epsilon>0$ there exist $N\in\mathbb{N}$ such that for all $n>N$ you have that $$\left|\frac{n}{2^n}-0\right|<\epsilon.$$
