Estimates involving sums with binomials When calculating some probabilities, I got sums of the form
$$\sum_{j=0}^c {j+a+b \choose a} p^j,$$ for integers $a, b, c > 0$.
Does someone know closed forms for these values?
 A: I don't have time right now to work out the exact answer, but the methods and identities in the middle third of this blog post will give a closed form for fixed $a$.  Use the geometric series formula to find $\sum_{j=0}^c p^{a+b+j}$, differentiate $a$ times with respect to $p$, then divide by $a! p^b$.
If a closed form for fixed $a$ isn't good enough, you should probably be more precise about which of your parameters are large and which are small.  
Edit:  Still don't have time to give a complete answer, but here's a fun trick.  Instead of computing the answer for fixed $a$ we can write down a generating function
$$\displaystyle P_{b,c}(x) = \sum_{a=0}^{\infty} x^a \sum_{j=0}^c {a+b+j \choose a} p^j$$
then exchange the order of summation, giving
$$\displaystyle \begin{align}
P_{b,c}(x) &= \sum_{j=0}^c p^j \sum_{a=0}^{\infty} {a+b+j \choose a} x^a \\
 &= \sum_{j=0}^c p^j \frac{1}{(1 - x)^{b+j+1}} \\
 &= \frac{1}{(1 - x)^{b+1}} \sum_{j=0}^c \left( \frac{p}{1-x} \right)^j \\
 &= \frac{1}{(1 - x)^{b+1}} \frac{1 - \left( \frac{p}{1-x} \right)^{c+1}}{1 - \frac{p}{1-x}} \\
 &= \frac{(1-x)^{c+1} - p^{c+1}}{(1 - x)^{b+c+1}(1 - p - x)}. \end{align}$$
Then the coefficient of $x^a$ of this rational function is the number you want.  Not sure how useful that is for you, but you might be able to extract a useful asymptotic from it.  The identity in the second line is the binomial theorem for negative exponents.
A: If you don't need an exact closed form or just care about the asymptotic growth Stirling's Approximation might be useful. I've found it useful in the past for generating good asymptotic bounds on similar functions.
