Compute $\lim_{n\to\infty} \frac {1}{n^k}\binom{n}{k}$

Question

Evaluate $$\lim_{n\to\infty} \frac {1}{n^k}\binom{n}{k} \,, \quad k\in\mathbb{Z}_+^*$$

What I tried so far:

I rewrote:

$$\lim_{n\to\infty} \frac {1}{n^k}\binom{n}{k}=\frac 1{k!}\lim_{n\to\infty}\frac {n(n-1)\cdots(n-k+1)}{n^k}=L$$

Applying $\ln$ to both sides we get:

$$\lim_{n\to\infty}\ln\left(\frac 1{k!}\right)+\sum_{i=1}^{k-1}\ln\left(\frac{ n-i}{ n^k}\right)$$

And thought maybe I could get a Riemann somewhere but it doesn't get me anywhere...

EDIT:

Solved it with Surb's hint the product of the limit is the limits of the product:

we get:

$$\frac 1{k!}\lim_{n\to\infty}\frac {n(n-1)\cdots(n-k+1)}{n^k}=\frac {1}{k!}\lim_{n\to\infty}\frac {n}{n}\lim_{n\to\infty}\frac{n-1}{n}\cdots\lim_{n\to\infty}\frac{n-k+1}{n}=\frac 1{k!}$$

Your first expression in "Tried ratio test " tends to $\frac{1}{k!}\not=0$ — Duchamp Gérard H. E., May 09 '18 at 16:26

score 2 · Accepted Answer · answered May 09 '18 at 16:19

2

Hint

The product doesn't depend on $n$. The limit of the product is therefore the product of the limit.

answered May 09 '18 at 16:19

Surb

55,662

What do you mean? – C. Cristi May 09 '18 at 16:20
It's n choose k, I mean that's what i wanted to be there... – C. Cristi May 09 '18 at 16:21
3

Indeed, but $$\frac{n(n-1)...(n-k+1)}{n^k}=\prod_{i=0}^{k-1}\frac{n-i}{n}$$ and as you can see, the product doesn't depend on $n$ (in the sense that when $n$ is bigger and bigger, you still have a product with $k$ elements). So the result is the product of the limits :-) (and easy to compute) – Surb May 09 '18 at 16:22
Oh, I see what you mean now – C. Cristi May 09 '18 at 16:23
So the limit is $\frac {1}{k!}$? – C. Cristi May 09 '18 at 16:24
Also what is wrong with ratio test? – C. Cristi May 09 '18 at 16:25
Yes, exactly :-) – Surb May 09 '18 at 16:25
1

Ratio test ? What is the context ? It's not clear... – Surb May 09 '18 at 16:26

score 1 · Answer 2 · answered May 09 '18 at 16:28

By ratio test

$$\frac {\begin{pmatrix}n+1\\k\end{pmatrix}}{(n+1)^k}\frac {n^k} {\begin{pmatrix}n\\k\end{pmatrix}}=\left(\frac{n}{n+1}\right)^k\frac{(n+1)!}{k!(n+1-k)!}\frac{k!(n-k)!}{n!}=\left(\frac{n}{n+1}\right)^k\frac{n+1}{n+1-k}\to 1$$

thus it is inconclusive.

Note that by binomial coefficient approximation

$${n \choose k} \sim \frac{n^k}{k!}$$

then

$$ \frac {\begin{pmatrix}n\\k\end{pmatrix}}{n^k}\sim \frac1 {k!} $$

Compute $\lim_{n\to\infty} \frac {1}{n^k}\binom{n}{k}$

2 Answers2