Is the sequence $\frac{e^n}{n}$ convergent? Is the sequence $$\frac{e^n}{n}$$ convergent?
I think it is not because $\log n < n$, implying that $\frac{e^n}{n} >1$ and hence the limit does not exist. Which probably also means that the sequence is unbounded.
Am I right? Please correct me if I am wrong.

Binomial expansion:
$e=1+x$ , where $x >0$, $n>1,$  positive integer.
$e^n = (1+x)^n = 1+nx+(n(n-1)/(2!))x^2 +....>$
$(n(n-1)/2)x^2$.
Hence 
$e^n/n \gt ((n-1)/2)x^2.$
Let $K >0$, real.
Archimedean principle:
There is a $N \in \mathbb{Z^+}$ such that 
$N >2K/(x^2)+1$.
For $n \ge N$:
$e^n/n \gt (n-1)x^2/2 \ge$
$(N-1)x^2/2 > K$, i.e.
$\lim_{n \rightarrow \infty} e^n/n = \infty.$
