I believe the $\lim_{n \to \infty} \frac{(3n)!(1/27)^n}{(n!)^3}$ -> 0.
But I am not sure if my reasoning is correct.
Because there is a higher power in the denomination that the numerator, the limit goes to 0?
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Maybe not strictly a duplicate, but closely related to: http://math.stackexchange.com/questions/1769570/find-lim-n-to-infty-left-frac33nn33n-right1-n – almagest May 03 '16 at 20:51
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yes, it goes to zero, because acctually you have $\lim_{n\rightarrow\infty}\dfrac{1}{(n!)^2(27)^{n-1}}$ – Fernando May 03 '16 at 20:52
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2@Fernando: careful... – parsiad May 03 '16 at 20:55
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You are right, I made terrible mistakes – Fernando May 03 '16 at 21:02
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Use Stirling's approximation $$ n!\sim \sqrt{2\pi n} (n/e)^n $$ for large $n$ to write $$ \frac{(3n)!(1/27)^n}{(n!)^3}\sim\frac{\sqrt{2\pi 3 n} (3n/e)^{3n} (1/27)^n }{(\sqrt{2\pi n} (n/e)^n)^3 }=\frac{\sqrt{3}}{2n\pi}\to 0 $$ for large $n$.
Pierpaolo Vivo
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For $n\geq 2$ we have $\int_{n-1}^n\log x; dx<$ $\log n< $ $\int_n^{n+1}\log x;dx.$ So $n(\log n)-n+1=$ $\int_1^n\log x;dx<$ $\log n!$ $<\int_2^{n+1}\log x; dx=$ $(n+1)\log (n+1)-n+1-2\log 2$ which is cruder than Stirling but ought to work for this Q. – DanielWainfleet May 03 '16 at 22:37
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$$\lim_{n\to \infty} \frac{(3n)! (1/27)^n}{(n!)^3} = \lim_{n\to \infty} \frac{(3n)!}{(3^n n!)^3}$$ Let's solve it with a ram. It's obvious that $1 < (3n)! < 3^{3n} * n^n $ and $n^n < n! n!$ so we can use the Squeeze theorem: $$\lim_{n\to \infty} \frac{1}{(3^n n!)^3} = 0; \lim_{n\to \infty} \frac{3^{3n} * n^n}{(3^{n} n!)^3} = \lim_{n\to \infty} \frac{n^n}{(n!)^3} = 0 $$
Kuroneko
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