$a_n = {{2^n}\over n}$ diverges? How do I formally show that the sequence $a_n = {{2^n}\over n}$ diverges using a $\delta$-$\epsilon$ argument?
 A: Here is a hint, since I don't know what your thoughts on the problem are. 
Pick an arbitrarily large $M>0$. Can you show that 
$$\frac{2^n}{n}\geq M$$
for every big enough $n$?
A: You may prove that, for $n\geq1$,
$$
2^n\geq n \times 1.2^n
$$ giving, for $n\geq1$,
$$
{{2^n}\over n}\geq 1.2^n
$$ then both sequences diverge as $n \to \infty$.
A: Consider the ratio of successive terms:
$$ \frac{2^{n+1}/(n+1)}{2^n/n} = 2\frac{n}{n+1}. $$
Since $2/3<n/(n+1)<1$ for $n>2$, the ratio between successive terms is always larger than $4/3$, say, so $a_n>(4/3)^{n-1}a_1$, which tends to $\infty$ since $4/3>1$. (If that's not clear enough, applying the increasing function $\log$ to both sides gives $\log{a_n}>n\log{(4/3)}+\log{a_1}$, and hence $\log{a_n} \to \infty$, &c.)
A: I am adding this answer just in case your difficulty is with the concept of "$\delta$-$\epsilon$ argument."  The use of deltas and epsilons is de rigueur when one is trying to prove that a limit of a function of a "continuous" variable is a particular value, for example a real function of a real variable tending to $1$ as its argument $x$ tends to $0$.  
When we speak of sequences $a_n$ converging to a particular value $A$ as $n\to\infty$, we use only epsilons:  the condition to be proved is that, for all $\epsilon>0$, we have $|a_n-A|<\epsilon$ for sufficiently large $n$, i.e., all $n\ge N$ where $N$ is dependent on $\epsilon$.  (If the sequences are not real or complex but in some more complicated space, $|a_n-A|$ gets replaced by something else.)  
When we speak of sequences $a_n$ diverging, we use neither epsilons nor deltas, but the argument could be called, I suppose, a $\delta$-$\epsilon$ argument.  Focus for concreteness on real sequences.  What we have to prove is that, given any $A\in \mathbb{R}$, $a_n\ge A$ for sufficiently large $n$, i.e., all $n\ge N$ where $N$ is dependent on $A$.  That is the case you are in.
As for how you get there, I think Charleston Crabb's answer is fine as a hint.  Consider alternatively that $${2^{n+1}\over n+1}=2{n \over n+1}{2^n\over n}\ge {4\over 3}{2^n\over n}$$ for $n\ge 2$, and consider that for $a>1$, for example for $a=4/3$, $a^n$ increases without bound.
A: You can see the order of growth of $2^n > n$ for all $n=1,2,...$. So, since order of growth of nominator is larger than order of growth of denominator it is obvious that sequence diverges. So,
$$\lim_{n \to \infty}a_n=\lim_{n \to \infty} \frac{2^n}{n}= \infty$$
