Consider $\displaystyle\sum_{k =n}^{5n} \binom{k-1}{n-1} \frac{1}{5^n} \cdot \left(\frac{4}{5}\right)^{k-n}$. We want to find it limit when $n$ goes to $+\infty$.
I've tried :
$$\displaystyle \lim_{n \to \infty} \sum_{k =n}^{5n} \binom{k-1}{n-1} \frac{1}{5^n}\cdot \left(\frac{4}{5}\right)^{k-n} = \lim_{n\to \infty} \left[\sum_{k = 0}^{\infty} \binom{n+k-1}{n-1}p^n (1-p)^k – \sum_{k=0}^{\infty} \binom{5n+k-1}{n-1} p^n (1-p)^{4n+k}\right]$$
Now the first term gives us :
$$\displaystyle \left(\frac{p}{1+(1-p)}\right)^n,$$ but what can we do with the second one?
As I understand it should be the probability problem (it look's like negative binomial distribution).
Best Answer
You are correct. It seems to be a probability problem. Namely the sum in question is the probability that one needs no more than $\frac np$ flips to obtain $n$ heads, $p$ being the probability of heads on a single flip. In your case $p=\frac15$.
Thus as $n\to\infty$ the sum tends to $\frac12$ as the pdf of the event "you need $k$ flips to obtain $n$ heads" tends to the normal distribution with the maximum at $k=\frac np$.
P.S. According to WA the sum for finite $n$ is expressed via hypergeometric function.