How to prove that, $n$ approach infinity and $0\leq k\leq n$ : $$ \lim_{n\to\infty} \dfrac{1}{k!}\times\dfrac{n!}{(n-k)!\times n^k} = \dfrac{1}{k!}$$
I have no idea to begin the proof...
Thanks in advance for help.
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You want an $\epsilon-\delta$ proof or just want to know how to calculate that limit? – Kushal Bhuyan Jun 12 '16 at 16:57
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The right side is not equal to $\binom{n}{k}$ by the way. ($n^k$ instead of $k!$) – Maximilian Gerhardt Jun 12 '16 at 17:00
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@MaximilianGerhardt I can´t follow your comment. – callculus42 Jun 12 '16 at 17:06
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I meant the title, but yeah, now I see that it is true if you multiply them together. Wrong comment. – Maximilian Gerhardt Jun 12 '16 at 17:09
3 Answers
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$$\frac{n!}{(n-k)!} = n(n-1)\cdots(n-(k+1))$$ is a polynomial in $n$ of degree $k$. So you have the limit towards infinity of a rational function whose numerator and denominator have the same degree. Thus the limit is the ratio of the leading coefficients.
$$\lim\limits_{n\to\infty} \frac{\left(\frac{n!}{(n-k)!}\right)}{n^k} = 1$$
Ken Duna
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It is not hard to prove. Here is a link that talks a bit about it: Link – Ken Duna Jun 12 '16 at 17:29
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Ok thanks, just factor out the leading coefficient and the remainder of polynomial over the coefficient tends to 1? – SuperFoxy Jun 12 '16 at 17:35
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For the limit used in this answer, see: Limits involing factorials $\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$ and other questions linked there. – Martin Sleziak Apr 13 '19 at 09:08
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Quite an overkill, but since $n!=\Gamma(n+1)$, your limit is a consequence of Gautschi's inequality.
Jack D'Aurizio
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