TITLE: Limit of a sum (no probabilities)
QUESTION [5 upvotes]: Show that $$\lim_{n\to+\infty}\left(\frac{2}{3}\right)^n\sum_{k=0}^{[n/3]}\binom{n}{k}2^{-k}=\frac{1}{2}$$ without using probabilities.
$[\;\cdot\;]$ denotes the integer part.

REPLY [2 votes]: One can also try contour integration (and it works).
Yet another idea, tailored specifically to "slices of the binomial formula", is to use $$\sum_{k=0}^{m}\binom{n}{k}a^{n-k}(1-a)^k=(n-m)\binom{n}{m}\int_0^a x^{n-m-1}(1-x)^m~dx\qquad(0\leqslant m<n)$$ (easily verified using induction on $m$, with the induction step being simply integration by parts).
The expression under the limit sign is obtained at $a=2/3$ and $m=\lfloor n/3\rfloor$. Writing $n=3m+r$, $$S_n:=\left(\frac23\right)^n\sum_{k=0}^{m}\binom{n}{k}2^{-k}=(n-m)\binom{n}{m}\int_0^{2/3}x^{r-1}\big(x^2(1-x)\big)^m dx.$$
Note that $x=2/3$ is the (local) maximum of $x^2(1-x)$. Substitute $x=\displaystyle\frac23\left(1-\frac{y}{\sqrt{3m}}\right)$: $$S_{3m+r}=\underbrace{\frac{2m+r}{m!\sqrt{3m}}\frac{(3m+r)!}{(2m+r)!}\frac{2^{2m+r}}{3^{3m+r}}}_{{}\to 1/\sqrt\pi\text{ when }m\to\infty}\int_0^{\sqrt{3m}}\left(1-\frac{y^2}{m}+\frac{2y^3}{(3m)^{3/2}}\right)^m\left(1-\frac{y}{\sqrt{3m}}\right)^{r-1}dy.$$
With $m\to\infty$, the integral tends to $\int_0^\infty e^{-y^2}dy=\sqrt\pi/2$, and the rest is done by Stirling's formula.