How to evaluate the limit $\lim_{n\to\infty}\left(\frac{\binom{n}{k} }{n^k}\right) $

Question

How do we evaluate the limit $$\lim_{n\to\infty}\left(\frac{\binom{n}{k} }{n^k}\right) $$

I know the sum of the combinations is $2 ^ k$, but the ratio of $2 ^ k$ and $n ^ k$ tends to $0$?

Answer 1

Since$$\binom nk=\frac{n(n-1)\ldots(n-(k-1))}{k!},$$you have$$\frac{\binom nk}{n^k}=\frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\ldots\left(1-\frac{k-1}n\right)$$and therefore your limit is equal to $\frac1{k!}$.

Answer 2

Hint: $$\binom{n}{k} = \frac{1}{k!}(n)(n-1)\cdots(n-k+1).$$

Answer 3

Hint. Note that for a fixed value of $k$, $$\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}=\frac{n^k+O(n^{k-1})}{k!}.$$