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.
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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$
Ishfaaq
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Your first inequality is wrong. You can simply replace $\frac1n$ by $0$. – Jul 21 '14 at 07:55
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@Sami Ben Romdhane: You're right. Edited. – Ishfaaq Jul 21 '14 at 07:59
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@Ishfaaq Thanks. How would you modify the first approach to show that $ne^{-an}$ tends to 0 where $a>0$? – user137090 Jul 21 '14 at 09:18
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1... Why not simply note that $e^{an}\geq a^2n^2/2$? – Jonas Dahlbæk Jul 21 '14 at 09:43
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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.
Jonas Dahlbæk
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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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Then you still need to show that $e^{-n}$ converges to zero, which is basically the same thing. – Jonas Dahlbæk Jul 21 '14 at 07:57
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1No 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. – Jul 21 '14 at 08:00
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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. – Jonas Dahlbæk Jul 21 '14 at 08:02
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What's more basic: the fact that $e^n\rightarrow \infty$ or that $e^n>\frac{n^2}2$? – Jul 21 '14 at 08:09
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1$e^n>n^2/2$ is a stronger result, but that does not make it any less basic. How would you prove $e^n\rightarrow\infty$? – Jonas Dahlbæk Jul 21 '14 at 08:13
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I remember when I first study this function that we defined it as the inverse function of the $\ln$ function which is the primitive of $\frac1x$ that vanishes on $1$. We proved that the derivative of $e^x$ is $e^x$ using the chain rule. We studied the function $e^x-x$ and we proved that $e^x>x$ for all $x\ge0$ so we found that $e^x\to\infty$ at $\infty$. All this are basic for me and at this time we did not even hear about the series. – Jul 21 '14 at 08:26
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1The 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. – Jonas Dahlbæk Jul 21 '14 at 08:37