Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the limit


share|cite|improve this question
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

2 Answers 2

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).

share|cite|improve this answer

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. $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.