Truncated binomial has limit?

Question

When $n$ goes to infinity through the odd numbers, $n=1,3,5,7,\dots$

$$\dfrac{\sum\limits_{k=0}^{\frac{n-1}{2}} \binom{n}{k}(1+(\frac{a}{n})^{0.5})^k}{(2+(\frac{a}{n})^{0.5})^n}$$

seems to have a limit. Why? And what is it, in terms of $a$?

Answer 1

Let us denote by $a_n$ the quantity in question

$$ a_n = \sum_{0 \leq k \leq n/2} \binom{n}{k} \frac{\left( 1 + \sqrt{\frac{a}{n}} \right)^k}{\left( 2 + \sqrt{\frac{a}{n}} \right)^n } $$

We may rewrite $a_n$ as

$$ a_n = \underbrace{ \left( \frac{1+\sqrt{\frac{a}{n}}}{1+\sqrt{\frac{a}{n}}+\frac{a}{4n}} \right)^{n/2} }_{ =: b_n } \sum_{0\leq k\leq n/2} \binom{n}{k}\frac{1}{2^n} \left( 1 + \sqrt{\frac{a}{n}} \right)^{k-n/2}. $$

Now let us write $S_n \sim \operatorname{Bin}(n,1/2)$ and write $S_n = \frac{n}{2}+\frac{\sqrt{n}}{2} Z_n$. The classical CLT tells that $Z_n \Rightarrow \mathcal{N}(0, 1)$. Then

$$ a_n = b_n \mathbb{E}\left[ \left( 1 + \sqrt{\frac{a}{n}} \right)^{\sqrt{n} Z_n/2} \mathbf{1}_{\{ Z_n \leq 0 \}} \right] $$

Taking $n\to\infty$ and combining with the fact that $b_n \to e^{-a/8}$, we have

$$ \lim_{n\to\infty} a_n = e^{-a/8} \int_{-\infty}^{0} \frac{1}{\sqrt{2\pi}} e^{\sqrt{a}z/2} e^{-z^2/2} \, dz. $$

The last integral can be written in terms of the error function. Indeed, it computes to

$$ \lim_{n\to\infty} a_n = \frac{1}{2}\operatorname{erfc}\left(\sqrt{\frac{a}{8}}\right) $$

where $\mathrm{erfc}$ is the complementary error function.