How calculate the limits as n->infinity of:$$(n^n)/n!$$ and $$(n^{n^{0.5}})/(2^n)$$, my attempts: the second have to do something with : $$(n^{(n^{0.5}))}=2^{\log(\log n)}$$......log with base 2. The first I guess goes to infinity by intuition...
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Stirling formula for 1st, use $a^b=\exp(b\log a)$ for 2nd part. – Milly Nov 01 '14 at 19:17
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The first question is a duplicate: Limit of factorial function: $\lim\limits_{n\to\infty}\frac{n^n}{n!}.$ – Hakim Nov 01 '14 at 19:17
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Well, the first blows up as $n^n$ eventually grows much faster than $n!$ – Daniel Goldman Nov 01 '14 at 19:18
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Hint: For the second, write $n^{n^{1/2}}= 2^{n^{1/2}\log_2 n}$. Then $$ \frac{n^{n^{1/2}}}{2^n} = \frac{2^{n^{1/2}\log_2 n}}{2^n} = 2^{\sqrt{n}\log_2 n - n}. $$ What can you say about the limit of $\sqrt{n}\log_2 n - n$?
For the first, either use Stirling's approximation; or if you cannot, the simpler fact that $$n^n = n^{\frac{n}{2}}\cdot n^{\frac{n}{2}}$$ while $$ n! = n(n-1)\dots\left(\frac{n}{2}+1\right)\cdot \left(\frac{n}{2}\right)! < n^{\frac{n}{2}}\cdot \left(\frac{n}{2}\right)! < n^{\frac{n}{2}}\cdot \left(\frac{n}{2}\right)^{\frac{n}{2}} $$ where the inequalities come from termwise comparisons in the products.
Clement C.
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Hum... I assumed you could take for granted the fact that $\frac{\log^b n}{n^a}\xrightarrow[n\to\infty]{} 0$, for any $a,b > 0$. – Clement C. Nov 01 '14 at 19:35
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@user188750 Can you find the bigger of $e^n$ and $n^2$? The required result follows. – user1001001 Nov 01 '14 at 19:37