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.
 A: Following your idea to use logarithm we have that
$$\log \left(\frac{e^n}{n}\right)=\log e^n-\log n=n-\log n=n\left(1-\frac{\log n}n\right)\to \infty\cdot (1-0) \implies \frac{e^n}{n}\to \infty$$
but we need to prove that $\frac{\log n}n\to 0$.
As an alternative it suffices to show, for example by induction, that
$$e^n>n^2$$
to immediately conclude by squeeze theorem that
$$\frac{e^n}{n}\ge \frac{n^2}{n}=n\to \infty$$

For the proof by induction we have


*

*base case: $n=2 \implies e^2>2^2=4$

*induction step: assume $e^n>n^2$ then


$$e^{n+1}=e\cdot e^n\stackrel{Ind.Hyp.}>e\cdot n^2\stackrel{?}>(n+1)^2$$
and the latter is true indeed
$$e\cdot n^2\stackrel{?}>(n+1)^2=n^2+2n+1$$
$$(e-1)n^2-2n-1\stackrel{?}>0$$
which is true for $n\ge 2$.
A: We want to check if $$L = \lim_{n\to\infty} \frac{e^n}{n}$$ is finite.
So,
$$\begin{align}L &= \lim_{n\to\infty} \frac{e^n}{n} \\
&= \lim_{n\to\infty}\frac{\frac{d(e^n)}{dn}}{\frac{dn}{dn}}\tag{L'Hopital's rule}  \\
&=\lim_{n\to\infty}e^n =\infty\end{align} $$
So, it is not convergent.
A: No it is not. Notice that $$a_{n+1}=\dfrac{e^{n+1}}{n+1}=e\cdot \dfrac{e^n}{n}\cdot \dfrac{n}{n+1}>\dfrac{e}{2}a_n>1.3a_n$$therefore $$a_{n+1}>1.3a_n>(1.3)^2a_{n-1}>\cdots>(1.3)^na_1=e\cdot(1.3)^n$$which is unbounded since $e\cdot(1.3)^n$ is unbounded.
A: 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.
$\lim_{n \rightarrow \infty} e^n/n = \infty.$
A: By the ratio test,$$\frac{a_{n+1}}{a_n}=\frac{e\,n}{n+1}>1.$$ 
A: The sequence is, obviously, divergent, but the fact You have provided cannot serve as a proof: there are many sequences $a_n,b_n$ such that $\frac{a_n}{b_n}>1$ and yet $\frac{a_n}{b_n}$ converges, take for instance $a_n=2, b_n=1$.
You can try to prove that the sequence is unbounded, divergence will follow trivially. There are several tricks that can be used: you can, for instance, apply de l'Hospital rule.
A: Note that
$$
\frac{a_{n+1}}{a_n}=\frac{en}{n+1}\to e>1
$$
as $n\to \infty$. In particular there exists $N$ such that $n\ge N$ implies that
$$
\frac{a_{n+1}}{a_n}>(e-1)>1
$$
i.e. for $n\geq N$, $a_{n+1}>a_N(e-1)^{n-N+1} \to \infty$ as $n\to \infty$ as desired.
A: By Otto Stolz theorem, $$\lim_{n \to \infty}\frac{e^n}{n}=\lim_{n \to \infty}\frac{e^n-e^{n-1}}{n-(n-1)}=\lim_{n \to \infty}e^{n-1}(e-1)=+\infty.$$
