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What is the limit


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Hint: the numerator is $e^{n\log n}$ – Thomas Andrews Oct 18 '12 at 19:40
You can then combine them, but that is not very helpful, since whatever I do I seem to end with an indeterminate form. – Edy Oct 18 '12 at 20:19
Different hint: $\left(\dfrac{n}{e^{n^{1/2}}}\right)^n$. Proving the inside part approaches $0$ is easier. And if $a_n\to0$, then $a_n^n\to0$ too. – alex.jordan Oct 18 '12 at 21:47
Come to think of it, isn't this a candidate sequence for the Root Test? The Root Test will imply these terms have a finite sum, which in turn proves the terms individually approach $0$. – alex.jordan Oct 18 '12 at 21:49

Observe that $n^n = e^{n\ln n}$. So the limit becomes $$\lim_{n \to \infty} \frac{e^{n\ln n}}{e^{n^{3/2}}}.$$ Without applying l'Hospital's Rule, which one do you think grows faster: $n\ln n$ or $n^{3/2}$? This is a shortcut (l'Hospital's rule in disguise).

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We will be using this result

Theorem: If ${a_n}$ be a sequence such that $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}= a\,,$ then

1) if $|a|<1$, then $\lim_{n\to \infty}a_n =0 \,,$

2) if $ a>1$, then $\lim_{n\to \infty}|a_n| =\infty \,.$

Another fact, we need to achieve our task is

$$ (n+1)^{\frac{3}{2}} = n^{\frac{3}{2}} + \frac{3}{2}\sqrt{n} + O(\frac{1}{\sqrt{n}})\,, $$

which can be derived by the binomial theorem. Now, let $ a_n = \frac{n^{n}}{e^{n^{3/2}}}\, $ then

$$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{n^{n+1} e^{n^{3/2}}}{ e^{(n+1)^{3/2} }n^n } = \lim_{n\to\infty} \frac{n e^{n^{3/2}}}{e^{n^{3/2}+\frac{3}{2}\sqrt{n}+O(1/\sqrt{n})}}= \lim_{n\to\infty} \frac{n}{e^{\frac{3}{2}\sqrt{n}+O(1/\sqrt{n})}}=0\,.$$

Since $ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0 \,,$ then by part $(a)$ of the theorem $\lim_{n\to\infty} a_n = 0. $

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