# Prove $ne^{-n}$ converges to zero

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

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 $\dfrac{2}{e} \lt 1$

• Your first inequality is wrong. You can simply replace $\frac1n$ by $0$.
– user63181
Jul 21, 2014 at 7:55
• @Sami Ben Romdhane: You're right. Edited. Jul 21, 2014 at 7:59
• @Ishfaaq Thanks. How would you modify the first approach to show that $ne^{-an}$ tends to 0 where $a>0$? Jul 21, 2014 at 9:18
• ... Why not simply note that $e^{an}\geq a^2n^2/2$? Jul 21, 2014 at 9:43

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

• so we get that $ne^{-n}{\leq}\frac{2}{n}$? Jul 21, 2014 at 12:50
• Yes, precisely so. Jul 21, 2014 at 21:01

We can use the L'Hôpital's rule to get the result easily:

$$\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac x{e^x}=\lim_{x\to\infty}\frac1{e^x}=0$$

• Then you still need to show that $e^{-n}$ converges to zero, which is basically the same thing. Jul 21, 2014 at 7:57
• No it isn't the same thing! The desired limit has the indeterminate form $\infty\times 0$ while $e^n$ tends to $\infty$ is a basic result.
– user63181
Jul 21, 2014 at 8:00
• I would argue that $e^n\rightarrow\infty$ is as basic a result as $e^n/n\rightarrow \infty$. I guess it is a matter of taste. Jul 21, 2014 at 8:02
• $e^n>n^2/2$ is a stronger result, but that does not make it any less basic. How would you prove $e^n\rightarrow\infty$? Jul 21, 2014 at 8:13
• The exact same ideas would lead you to consider the function $e^x-x^2/2$ and conclude that $e^x\geq x^2/2$ for all $x\geq 0$. This is what I mean by the result being equally basic, it is just a different way to make use of the same ideas. Jul 21, 2014 at 8:37