Prove $ne^{-n}$ converges to zero
By Sarah Richards
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
$\endgroup$ 03 Answers
$\begingroup$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$
$\endgroup$ 4 $\begingroup$Consider the fact that $e^n\geq n^2/2$. The inequality is a simple consequence of the series expansion of the exponential function.
$\endgroup$ 2 $\begingroup$We can use the L'Hôpital's rule to get the result easily:
$$\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac x{e^x}=\lim_{x\to\infty}\frac1{e^x}=0$$
$\endgroup$ 7
