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.




$$ 0 \le \frac{n}{e^n} \le \frac{2^n}{e^n} = \left({\frac{2}{e}} \right)^n $$

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$

  • $\begingroup$ Your first inequality is wrong. You can simply replace $\frac1n$ by $0$. $\endgroup$ – user63181 Jul 21 '14 at 7:55
  • $\begingroup$ @Sami Ben Romdhane: You're right. Edited. $\endgroup$ – Ishfaaq Jul 21 '14 at 7:59
  • $\begingroup$ @Ishfaaq Thanks. How would you modify the first approach to show that $ne^{-an}$ tends to 0 where $a>0$? $\endgroup$ – user137090 Jul 21 '14 at 9:18
  • 1
    $\begingroup$ ... Why not simply note that $e^{an}\geq a^2n^2/2$? $\endgroup$ – Jonas Dahlbæk Jul 21 '14 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.

  • $\begingroup$ so we get that $ne^{-n}{\leq}\frac{2}{n}$? $\endgroup$ – user137090 Jul 21 '14 at 12:50
  • $\begingroup$ Yes, precisely so. $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 21:01

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

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

  • $\begingroup$ Then you still need to show that $e^{-n}$ converges to zero, which is basically the same thing. $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 7:57
  • 1
    $\begingroup$ 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. $\endgroup$ – user63181 Jul 21 '14 at 8:00
  • $\begingroup$ 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. $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 8:02
  • 1
    $\begingroup$ $e^n>n^2/2$ is a stronger result, but that does not make it any less basic. How would you prove $e^n\rightarrow\infty$? $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 8:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.