prove that $(\frac{1}{6})^{4}\cdot\lim_{n\rightarrow\infty}\sum_{i=4}^{n}\binom{i-1}{3}(\frac{5}{6})^{i-4}=1$ I have to prove the following:
$(\frac{1}{6})^{4}\cdot\lim_{n\rightarrow\infty}\sum_{i=4}^{n}\binom{i-1}{3}(\frac{5}{6})^{i-4}=1$
any ideas?
thanks
 A: The argument below uses a  probabilistic interpretation.
Roll a fair die repeatedly until the fourth time we get a $1$. Let random variable $X$ be the number of rolls required. Then $X=i$  if we get three $1$'s in the first $i-1$ rolls, and a $1$ on the $i$-th roll. This has probability
$$\binom{i-1}{3}(1/6)^3 (5/6)^{i-1-3}(1/6).\tag{1}$$
We did this by direct computation, but for the general formula please look under negative binomial distribution.
The sum of the probabilities (1), from $i=4$ to $\infty$, is $1$, for with  probabiity $1$ the fourth $1$ will come up at some time. Thus 
$$\frac{1}{6^4}\sum_{i=4}^\infty \binom{i-1}{3}(5/6)^{i-4}=1.$$
That implies the desired result.
A: Hint :
$$\sum_{j=0}^\infty x^j =\frac{1}{1-x}$$ for $|x|<1$
$$\sum_{j=0}^\infty jx^j = \frac{x}{(1-x)^2}$$
for $|x|<1$
$$\sum_{j=0}^\infty j^2x^j =\frac{x(x+1)}{(1-x)^3}$$
for $|x|<1$
A: HINT:
Calculate the third derivative of the geometric series $\sum_{i=1}^{\infty}x^{i-1}$ as
$$\frac{d^3}{dx^3}\sum_{i=1}^{\infty}x^{i-1}$$
and evaluate at $x=5/6$ and you'll almost be done.
