TITLE: Find limit $\lim_{n\to\infty}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^{n}\left(\frac{4}{5}\right)^{k-n}$
QUESTION [9 upvotes]: The question is pretty straightforward.

Find $$\lim_{n\to\infty}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^{n}\left(\frac{4}{5}\right)^{k-n}$$

Attempt

Let's denote $F$ as 
$$
\begin{align*}
& \lim_{n\to\infty}\sum_{k=n}^{5n}{k-1 \choose n-1}\left(\frac{1}{5}\right)^{n}\left(\frac{4}{5}\right)^{k-n} = \\ 
&= \lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=n}^{5n}\left(\frac{4}{5}\right)^{k-n}\frac{(k-1)!}{(k-n)!} =  \\ &= \lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=0}^{4n}\left(\frac{4}{5}\right)^{k}\frac{(k+n-1)!}{k!} = F
\end{align*}
$$
Try to establish a lower and upper bounds of denoted function $F$, lower bound set to 0, and upper bound to $$\lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=0}^{\infty}\left(\frac{4}{5}\right)^{k}\frac{(k+n-1)!}{k!}$$
The sum inside of a limit equals to $(1 + \frac{4}{5})^{n}$ (by using Taylor series). By substituting this into the obtained upper bound, we're getting $0$ as an answer.

REPLY [4 votes]: Imagine an infinite sequence of independent trials with probability $1/5$ if success on each trial.
On average, then it takes $5$ trials to get one success.
Let $X$ be the number of trials needed to get $n$ successes.
Then the expected value of $X$ is $5n.$
The variance of $X$ is $20n.$
That last is somewhat more work to prove, but it is $n$ times the variance of the number of trials needed to get one success, since $X$ is the sum of $n$ independent copies of that random variable.
Thus the random variable $\dfrac{X-5n}{\sqrt{20n}}$ has expected value $0$ and standard deviation $1.$ And as $n\to\infty,$ the distribution of this random variable approaches that of the standard normal random variable $Z.$
$$
\frac{n-5n}{\sqrt{20n}} \le \frac{X-5n}{\sqrt{20n}} \le \frac{5n-5n}{\sqrt{20n}} = 0.
$$
So the limit is $\Pr(Z\le 0) = \dfrac 1 2.$