How can we find limit of the sequence $\dfrac{n!e^n}{n^n} $? I know its limit is infinity. First, I showed it is strictly increasing(by taking ratio of terms) and then showed it is not Cauchy sequence. I want to prove this by some other method, like comparing with other sequence. Can any one give some hint?
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By Stiriling's Formula the limit is $\infty$. – Kavi Rama Murthy Aug 06 '20 at 07:36
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How could you show that it is not Cauchy ? – Aug 06 '20 at 07:39
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2Also: Elementary proof for $\lim\limits_{n \to\infty}\frac{n!e^n}{n^n} = +\infty$ and Limit $c^n n!/n^n$ as $n$ goes to infinity and Is the sequence $(n!e^{n}/n^{n})$ convergent? – all found with Approach0 – Martin R Aug 06 '20 at 08:04
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One quick solution is using Stirling's approximation, which states
$n! \sim \sqrt{2 \pi n} \frac{n^n}{e^n}$
Then $\frac{n! e^n}{n^n} \sim \sqrt{2 \pi n}$ which approaches $\infty$ as $n$ approaches $\infty$.
Doctor Who
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Let me write little more exact estimation, from which, by the way, comes Stirling. $$\sqrt{2\pi n}\left(\frac{n}{e} \right)^n<n!$$
zkutch
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