How can I prove this limits result? When I did exercises in probability theory I found this limits as follows and verified it with Mathematica 8.0, and also noticed when $p=\dfrac12$ it shows that $\displaystyle p^n\sum_{k=0}^{n-1}\binom{k+n-1}{k}q^k \equiv \frac12$, but how it works?

$$
    \lim_{n\to\infty}p^n\sum_{k=0}^{n-1}\binom{k+n-1}{k}q^k=\begin{cases}0, & p<0.5\\ 0.5, & p=0.5\\ 1, & p>0.5\end{cases} \qquad p,q>0, p+q=1, n\in\mathbf N^*
$$

 A: If $p = \frac{1}{2}$ then $p = q$. By letting $j = n -1$ the limit becomes 
\begin{equation}
\lim_{j \to \infty} p^{j+1} \sum_{k = 0}^j {j + k \choose k} q^k = \lim_{j \to \infty} p^{1} \sum_{k = 0}^j {j + k \choose k} \frac{1}{2^{j + k}} 
\end{equation}
Here is the interesting part, if $q = \frac{1}{2}$ then the summation is always 1(I found this by taking partial sums). I havent had the time to prove the property, though I think a proof by induction can be used to show the summation is always 1. By knowing this the limit becomes
\begin{equation}
\lim_{j \to \infty} p^1 = p = \frac{1}{2};
\end{equation}
In fact I would be willing to bet that the expression is always equal to $\frac{1}{2}$ regardless of $j$ as long as $p = \frac{1}{2}$. Ill will post a proof of the property if I get one.
A: Player A has prob.$ p$ of winning any one game.Player B has prob. $q=1-p$ of winning any one game.A series of "best $n-1$ out of $2n-3$ games" is played. The series stops when one player has reached $n-1$ wins.The odds that player A wins the series is $$f(A,p,n-1)=\sum_{k=0}^{n-2}p^{n-1}q^k \binom {k+n-1}{k}.$$.Now if $p=q=1/2$ we have $f(A,p,n-1)=f(A,1/2,n-1)=1/2$ because B is just as likely to win the series.To evaluate the formula in the Q, with $p=q=1/2$,  we need to know, as well, that $$\lim_{n\to \infty}(pq)^{n-1}\binom {2n-1}{n-1}=0$$ for any $p\in [0,1]\wedge q=1-p.$................. Now,for $p>1/2$  would you think $\lim_{n\to \infty}f(A,p,n-1)$ could be less than $1$?
