I'm trying to compute the limit:
$$\lim_{n\to\infty} \frac{n!\cdot(3n)!}{(4n)!}$$
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Your link doesn't work. Try typing it out even if the formatting is poor. – Frank Mar 25 '14 at 21:45
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5xum already do it,thank you! – user2976686 Mar 25 '14 at 21:48
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Unfortunately he retained an $x$ in the limit! – Frank Mar 25 '14 at 21:51
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@Frank Indeed I did, on purpose. I edited the post to make it readable, but left the contents undisturbed for the OP to comment. He can fix the $n$ to $x$ easily. – 5xum Mar 25 '14 at 21:53
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OK, I understand - I don't know anything about the etiquette of editing I'm afraid! – Frank Mar 25 '14 at 21:54
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Related : https://math.stackexchange.com/questions/1464006/what-is-the-limit-of-n-cdot-2n-3n – Arnaud D. Oct 13 '20 at 09:04
2 Answers
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$$\frac{n!(3n)!}{4n!} = \frac{n(n-1) ... 1}{4n(4n-1)...(3n+1) } \leq \frac{1}{3n+1} \rightarrow 0 \quad \textrm{as} \quad n \rightarrow \infty. $$
Frank
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That inequality needs some justification, but the idea is quite alright I think. – 5xum Mar 25 '14 at 21:52
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2The only justification required is that $n \leq 4n, n-1 \leq 4n-1,....$, or equivalently $1 \leq 4$... – Frank Mar 25 '14 at 21:54
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@Frank Yes, it was not a criticism, I think your answer is very good. I just commented so OP would not think that he can write the inequality without thought. – 5xum Mar 25 '14 at 21:57
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*Hint:*$$\frac{(3n)!}{(4n)!} = \frac{1\cdot 2 \cdot 3\cdots 3n}{1\cdot 2\cdot 3 \cdots 3n\cdot (3n+1) \cdot (3n+2) \cdots (4n-1)\cdot 4n}=\frac{1}{(3n+1)(3n+2)\cdots 4n}$$
5xum
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hi,thanks but what about: $$\lim_{x\to\infty} \frac{n!}{(3n+1)...4n}$$ what i need to do now? – user2976686 Mar 25 '14 at 21:54
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Well, my answer is just a hint, so this is where you should stare at the equation for a while and see what comes up. In this case, both the denominator and numerator have $n$ factors. Try writing the fraction as a product of $n$ fractions and see what you can discover about each. The first is $\frac1{3n+1}$ which limits to $0$, if the others are bounded, you are done! – 5xum Mar 25 '14 at 21:56
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whan i tried to continue: $$\lim_{x\to\infty} \frac{1}{(4n)\cdot(3n+1)\cdot....\cdot1}$$
but this is limit to 0 no to inf,where is my mistake?
– user2976686 Mar 25 '14 at 22:11 -
1@5xum, when setting $n=1$ it goes to $\frac 1 {3(1)+1}=\frac 1 4$. Then $n=2 \mapsto \frac 2{7\cdot 8}$, $n=3 \mapsto \frac 6 {10\cdot 11\cdot 12 }$, et cetera. The denominator appears to be increasing faster than the numerator. – Graham Kemp Mar 25 '14 at 22:33
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