Can you explain me how to solve $ \lim_{n\rightarrow +\infty} \frac{ n!}{ ((n-k)!.n^k)}$
Asked
Active
Viewed 75 times
1
-
Do you mean: $$\lim_{n\to \infty} \frac{n!}{(n-k)!\cdot n^k}$$ – projectilemotion Mar 25 '17 at 00:44
-
Stirling approximation will do. – Simply Beautiful Art Mar 25 '17 at 00:44
-
http://math.stackexchange.com/questions/527002/why-does-displaystyle-lim-n-to-infty-fracnn-knk-equal-1/527003#527003 – lab bhattacharjee Mar 28 '17 at 08:31
2 Answers
2
Stirling's formula isn't required: $$\frac{n!}{(n-k)!}=n(n-1)\dotsm(n-k+1)\sim_\infty n^k, \enspace\text{so}\quad\frac{n!}{(n-k)!\,n^k}\sim_\infty\frac{n^k}{n^k}=1. $$
Bernard
- 179,256
-
that's right. I got a little bit confused to see if the decomposition of the n! would be k times or k+1 times – Vitor Aguiar Mar 25 '17 at 00:56
1
$$\dfrac{n!}{(n-k)!\cdot n^k}=\dfrac{\prod_{r=0}^{k-1}(n-r)}{n^k}=\prod_{r=0}^{k-1}\left(1-\dfrac rn\right)$$
lab bhattacharjee
- 279,016