Evaluate $\lim_{n \to \infty} \sum_{j=0}^{n}{{j+n-1} \choose j}\frac{1}{2^{j+n}}$ 
Evaluate $$\lim_{n \to \infty} \sum_{j=0}^{n}{{j+n-1} \choose j}\frac{1}{2^{j+n}}$$

I don't understand where to start.  Please help.
 A: As guy notes, the sum has a probabilistic interpretation.
Suppose that you  toss a fair coin until you have either $n$ heads 
or $n$ tails. The probability that you need $j+n$ tosses in total is 
$2{j+n-1\choose j}(1/2)^{j+n}$ for $0\leq j\leq n-1$. 
The factor of $2$ in the front comes from the two cases, whether the series ends with a head or a tail. 
Since the probabilities
add up to one, we have
$$2\sum_{j=0}^{n-1} {j+n-1\choose j}\left({1\over2}\right)^{j+n}=1,$$
and so
$$\sum_{j=0}^{n-1} {j+n-1\choose j}\left({1\over2}\right)^{j+n}={1\over2}.$$
The sum considered by the OP has one additional term that goes to zero. 
So as $n\to\infty$, 
$$\sum_{j=0}^{n}{{j+n-1} \choose j}\frac{1}{2^{j+n}}={1\over2}+{2n-1\choose n-1}{1\over 2^{2n}}\to{1\over2}.$$

Added: The additional term has the product representation
$${2n-1\choose n-1}{1\over 2^{2n}}={1\over2}\prod_{j=1}^n \left(1-{1\over 2j}\right).$$
Using the elementary arguments from my answer here we get the upper bound
$${2n-1\choose n-1}{1\over 2^{2n}}\leq{3\over 8\sqrt{n+1}},$$
which shows that the extra term goes to zero. Of course, there are other ways to do this.

See this question as well.
