TITLE: Prove $ne^{-n}$ converges to zero
QUESTION [5 upvotes]: How would I prove that $ne^{-n}$ converges to zero? I've tried $ne^{-n}<{\epsilon}$ and then logging both sides but no further progress could be made.
Thanks

REPLY [9 votes]: Hint
$$ 0 \le \frac{n}{e^n} \le \frac{2^n}{e^n} = \left({\frac{2}{e}} \right)^n  $$
We know that $ e \gt 2$ and hence the geometric series $\sum  \left({\frac{2}{e}} \right)^n$ converges which necessitates that $ \lim  \left({\frac{2}{e}} \right)^n = 0$.  Now we apply the Squeeze Theorem. 

You can use your approach too.
Let $\epsilon \gt 0$ be arbitrary. 
$$ \left|{\frac{n}{e^n}}\right|  =  \frac{n}{e^n} \le \frac{2^n}{e^n}  $$
Now, notice that $ \dfrac{2^n}{e^n} \lt \epsilon \iff \ln {\dfrac{2^n}{e^n}} \lt \ln \epsilon \iff n \ln \dfrac{2}{e}  \lt \ln \epsilon \iff n \gt \dfrac{\ln \epsilon}{\ln \dfrac{2}{e}  } $
where $\ln \dfrac{2}{e} \lt 0 $ since $ \dfrac{2}{e} \lt 1$

REPLY [7 votes]: Consider the fact that $e^n\geq n^2/2$. The inequality is a simple consequence of the series expansion of the exponential function.

REPLY [6 votes]: We can use the L'Hôpital's rule to get the result easily:
$$\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac x{e^x}=\lim_{x\to\infty}\frac1{e^x}=0$$