What function results from evaluating my function in the limit that n goes to positive infinity.

Question

I am looking at a special case of the binomial PDF, where $$p = \frac{1}{n}$$ i.e., $$f(n,k) = \frac{n!}{k!(n-k)!} \left(\frac{1}{n}\right)^k \left(1-\frac{1}{n}\right)^{n-k}$$

I have observed empirically that in the limit that $n\rightarrow \infty$, the two-variable PDF appears to become a new one-variable PDF: $f(n,k)\rightarrow g(k)$.

I would like a closed form expression for this new function $g(k)$, if such an expression exists.

Answer 1

You are looking to find $$\lim_{n\to\infty}\binom nk\left(\frac1n\right)^k\left(1-\frac1n\right)^{n-k}.$$ For $n\to\infty$, $$\binom nk=\frac{n^k}{k!}(1-o(1)),$$ since $\binom nk$ is a polynomial in $n$ with degree $n!$ and leading coefficient $1/k!$. Similarly, $$\left(1-\frac1n\right)^k=1-o(1);$$ also, $\lim_{n\to\infty}(1-1/n)^n=e^{-1}$. This means that the factors $$\left[\binom nk\left(\frac1n\right)^k\right]\left[\left(1-\frac1n\right)^n\right]\left[\left(1-\frac1n\right)^k\right]^{-1}$$ each have limits as $n\to\infty$ separately, and so the eventual limit is their product, $$\frac1{k!e}.$$

Answer 2

For your curiosity.

You can have more than the limit itself.

Rewriting $$f(n,k) =\frac{n!}{k!(n-k)!} \left(\frac{1}{n}\right)^k \left(1-\frac{1}{n}\right)^{n-k}=\frac1{k!}\color{blue}{\frac{ n!\,n^{-n}(n-1)^{n-k}}{(n-k)!}}$$ Take the logarithm of the blue piece and use Stirling approximation and continue with Taylor series to get $$\log\Bigg[\color{blue}{\frac{ n!\,n^{-n}(n-1)^{n-k}}{(n-k)!}} \Bigg]=-1+\frac{-k^2+3 k-1}{2 n}+\frac{-2 k^3+3 k^2+5 k-4}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using $x=e^{\log(x)}$ andcontinuing with Taylor series, therefore $$f(n,k)=\frac 1{e\,k!}\Bigg[1-\frac{k^2-3 k+1}{2 n}+\frac{3 k^4-22 k^3+39 k^2-8 k-5}{24 n^2}+O\left(\frac{1}{n^3}\right) \Bigg]$$