Limit of a sum (no probabilities) 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.
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The Question:
  $\ds{\lim_{n \to \infty}\bracks{\pars{2 \over 3}^{n}\
\sum_{k = 0}^{\left\lfloor n/3\right\rfloor}{n \choose k}2^{-k}} = \half}$

Thanks to the user @akotronis, I already read the 
Omar Kouba rigorous solution ( page 26 of Asymmetry Journal ).
However, I would like to show a solution which can be considered more or less heuristic. Namely:
For 'large' $n$, we note that the function $\ds{{n \choose z}2^{-z}}$ is 'strongly peaked around' $\ds{z = {n \over 3}}$ such that
$$
{n \choose z}2^{-z} \sim {n \choose n/3}2^{-n/3}\exp\pars{-\,{9 \over 4n}\bracks{z - {n \over 3}}^{2}}\quad\mbox{for 'large'}\ n
$$
and
\begin{align}
&\color{#f00}{\lim_{n \to \infty}\bracks{\pars{2 \over 3}^{n}\
\sum_{k = 0}^{\left\lfloor n/3\right\rfloor}{n \choose k}2^{-k}}} &
\\[3mm] & =
\lim_{n \to\infty}\braces{\pars{2 \over 3}^{n}{n \choose n/3}2^{-n/3}
\int_{0}^{\left\lfloor n/3\right\rfloor}
\exp\pars{-\,{9 \over 4n}\bracks{z - {n \over 3}}^{2}}\,\dd z}
= \color{#f00}{\half}
\end{align}

Note that the above integral can be expressed in terms of the Error Function
  $\mathrm{erf}$.

A: 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.
