$a_n = {{2^n}\over n}$ diverges? How do I formally show that the sequence $a_n = {{2^n}\over n}$ diverges using a $\delta$-$\epsilon$ argument?
Here is a hint, since I don't know what your thoughts on the problem are. 

Pick an arbitrarily large $M&gt;0$. Can you show that 
$$\frac{2^n}{n}\geq M$$
for every big enough $n$?
You can see the order of growth of $2^n &gt; n$ 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,

$$\lim_{n \to \infty}a_n=\lim_{n \to \infty} \frac{2^n}{n}= \infty$$