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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

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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$

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  • $\begingroup$ 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
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    $\begingroup$ ... Why not simply note that $e^{an}\geq a^2n^2/2$? $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 9:43
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Consider the fact that $e^n\geq n^2/2$. The inequality is a simple consequence of the series expansion of the exponential function.

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  • $\begingroup$ 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
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We can use the L'Hôpital's rule to get the result easily:

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ $e^n>n^2/2$ 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
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    $\begingroup$ 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

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