TITLE: Find the limit $\lim\limits_{n \to +\infty}\sum\limits_{k=n}^{3n} \binom{k-1}{n-1} \left(\frac{1}{3}\right)^n \left(\frac{2}{3}\right)^{k-n}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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 $\Pr(X \leq 0) = \frac12$.