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This question already has an answer here:

Why is $\lim\limits_{n\to\infty}\frac{3e^n}{n!}=0$?

$e^n=e\cdot e\cdots e$, $n$-times. I tried to find a constant $C>0$ and a $N\in \mathbb{N}$ such that $\frac{3e^n}{n!}\le C\frac{e}{n}$ for $n\ge N$, to apply the squeezing lemma afterwards. But I don't know how to estimate the fraction $\frac{3e^n}{n!}$, how can I estimate $\frac{3e^n}{n!}$ from above? Thank you.

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marked as duplicate by Sil, Foobaz John, Guy Fsone, Math Lover, Lord Shark the Unknown Dec 23 '17 at 22:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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\begin{align*} \dfrac{e^{n}}{n!}&\leq\dfrac{e}{1}\cdot\dfrac{e}{2}\cdot\dfrac{e}{3}\cdots\dfrac{e}{3},~~~~n\geq 3\\ &=\dfrac{e^{2}}{2}\cdot\left(\dfrac{e}{3}\right)^{n-2}\\ &\rightarrow 0 \end{align*}

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  • $\begingroup$ And there is nothing special about $e$. $\endgroup$ – Mark Viola Dec 23 '17 at 22:17
  • $\begingroup$ Yes, then choose large $n\geq a$ for any $a>1$ and do it in this fashion. $\endgroup$ – user284331 Dec 23 '17 at 22:18
  • $\begingroup$ thank you very much, it answeres my question $\endgroup$ – user472520 Dec 23 '17 at 22:24
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Just as an alternative solution: the series $$\sum_{n\geq 0}\frac{x^n}{n!}$$converges absolutely for any $x\in\Bbb R$ (as easily seen by the ratio test). In particular, it converges when $x=e$. A necessary condition for series convergence is that the sequence being summed tends to zero as $n\to\infty$.

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  • $\begingroup$ this is an awesome solution, thank you! $\endgroup$ – user472520 Dec 23 '17 at 22:24
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Hint1 the series $$\sum\frac{e^n}{n!}=e^e$$ converges

Hint2 Use Stirling formula Stirling's formula: proof?

$$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$

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  • $\begingroup$ thank you. Nice to see this alternative approach. $\endgroup$ – user472520 Dec 23 '17 at 22:25
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First of all, it's obvious that $\forall n \in \mathbb{N}, u_n = \frac{3e^n}{n!}\geq 0$.

Now let's study its behaviour: $$\frac{u_{n+1}}{u_n}=\frac{e^{n+1}}{(n+1)!}\frac{n!}{e^n}=\frac{e^1}{n+1}$$, thus the sequence is decreasing, minored by $0$, thus its limit is $0$.

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    $\begingroup$ thank you. Nice to see this alternative solution $\endgroup$ – user472520 Dec 23 '17 at 22:25
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Every time you increase $n$, you multiply the previous term by $e$ and divide it by the new $n$. Obviously the factors are very soon smaller than $1$.

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