Find the limit $\lim\limits_{n \to +\infty}\sum\limits_{k=n}^{3n} \binom{k-1}{n-1} \left(\frac{1}{3}\right)^n \left(\frac{2}{3}\right)^{k-n}$ The problem is to find the following limit:
$$\lim_{n \to +\infty}\sum\limits_{k=n}^{3n} \binom{k-1}{n-1} \left(\dfrac{1}{3}\right)^n \left(\dfrac{2}{3}\right)^{k-n}$$
I see that it looks similarly to the formula from the binomial theorem, but don't get how we can make use of it.
Any ideas would be greatly appreciated.
 A: The sum over all integers is
$$
\begin{align}
\sum_{k=n}^\infty\binom{k-1}{n-1}\left(\frac13\right)^n\left(\frac23\right)^{k-n}
&=\left(\frac13\right)^n\sum_{k=n}^\infty(-1)^{k-n}\binom{-n}{k-n}\left(\frac23\right)^{k-n}\\
&=\left(\frac13\right)^n\sum_{k=0}^\infty\binom{-n}{k}\left(-\frac23\right)^k\\
&=\left(\frac13\right)^n\left(\frac13\right)^{-n}\\[9pt]
&=1\tag{1}
\end{align}
$$
Furthermore,
$$
\frac{\binom{k}{n-1}\left(\frac13\right)^n\left(\frac23\right)^{k-n+1}}{\binom{k-1}{n-1}\left(\frac13\right)^n\left(\frac23\right)^{k-n}}=\frac23\frac{k}{k-n+1}\tag{2}
$$
which is $1$ when $k\approx3n$. That is, the summand is at its maximum near $k\approx3n$. By the method of Laplace, the sum to the maximum tends to one half the total sum, so we get that
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sum_{k=n}^{3n}\binom{k-1}{n-1}\left(\frac13\right)^n\left(\frac23\right)^{k-n}=\frac12}\tag{3}
$$

Application of Laplace's Method
Let
$$
f_n(k)=\binom{k-1}{n-1}\left(\frac13\right)^n\left(\frac23\right)^{k-n}\tag{4}
$$
From $(3)$, we can derive that for $k-3n=O\!\left(\sqrt{n}\right)$,
$$
\begin{align}
\log(f_n(k+1))-\log(f_n(k))
&=\log\left(\frac23\frac{k}{k-n+1}\right)\\
&=\log\left(\frac{1+\frac{k-3n}{3n}}{1+\frac1{6n}+\frac{k-3n}{2n}}\right)\\
&=\log\left(1-\frac{k-3n}{6n}\right)+O\!\left(\frac1n\right)\\
&=-\frac{k-3n}{6n}+O\!\left(\frac1n\right)\tag{5}
\end{align}
$$
From $(5)$, we get that
$$
f_n(k)=c\,e^{-\frac{(k-3n)^2}{12n}+O\left(\frac1{\sqrt{n}}\right)}\tag{6}
$$
Since the sum for all $k$ of $f_n(k)$ is $1$, we get
$$
f_n(k)=\frac1{\sqrt{12\pi n}}\,e^{-\frac{(k-3n)^2}{12n}}+O\!\left(\frac1n\right)\tag{7}
$$
Since $(7)$ is symmetric about $k=3n$, the sum for $k\lt3n$ will be approximately the same as the sum for $k\gt3n$. Thus, the sum for $k\lt3n$ will be approximately $\frac12$.
A: Consider $\{\xi_i, i \in \mathbb{N}\}$ -- iid $\mathrm{Geom}\left(\frac13\right)$ variables ($\xi_i \geq 1$).
$$
\Pr\left(\sum\limits_{i = 1}^n \xi_i \leq 3n \right) = \sum\limits_{k = n}^{3n}\Pr\left(\sum\limits_{i = 1}^n \xi_i = k \right) = \sum\limits_{k = n}^{3n} \binom{k - 1}{n - 1}\left(\frac13\right)^n \left(\frac23\right)^{k - n} 
$$
It is so because we iterate from $n$ to $3n$ (these are the all possible variants for the sum of $n$ geometric variables). Then there are $\binom{k - 1}{n - 1}$ variants to have the needed sum (stars and bars method) and the probability of each variant is really obvious because all of the variables have $1/3$ in pmf and other sum in exponent is overall $k - n$.
So, our answer is (remember that $\mathsf{E}\xi_i = 3$)
$$
\lim\limits_{n \to +\infty} \Pr\left(\sum\limits_{i = 1}^n \xi_i \leq 3n \right) = \lim\limits_{n \to +\infty} \Pr\left(\sum\limits_{i = 1}^n (\xi_i - \mathsf{E}\xi_i) \leq 0 \right) = \lim\limits_{n \to +\infty}\Pr\left(\frac{\sum\limits_{i = 1}^n(\xi_i - \mathsf{E}\xi_i)}{\sqrt{n\mathsf{D}\xi_1}} \leq 0 \right)
$$
The equation under the probability converges in distribution (due to central limit theorem) to $X \sim \mathcal{N}(0, 1)$ and obviously the limit is $1/2$ because $\Pr(X \leq 0) = \frac12$.
