# Bounding a hypergeometric sum

Is there a general method for finding upper bounds of hypergeometric sums?

The sum in particular I am trying to bound is $\sum_{j=1}^{k}\binom{k-1}{j-1}\binom{n-k-1}{j-1}4^{k-j}$.

This is a hypergeometric sum with ratio $\frac{(k-j)(n-k-j)}{4(j-1)^2}$.

Thanks.

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Here gross bound *Cauchy-Schwarz *\sum_{j=1}^{n} a_j^2 <= (\sum_{j=1}^{n} a_j)^2 when a_j >1 *\sum_{j=1}^{n} a_j <= \sum_{j=1}^N a_j for a_j>=0 and N>n *\sum_{j=1}^{k-1} { k-1 \choose j-1 } = 2^{k-1} using those you can get a (terrible) bound of 4^{n+k} (i think) –  Taylor Dupuy Jun 17 '11 at 8:54
those *'s are supposed to be new lines. –  Taylor Dupuy Jun 17 '11 at 8:55
Write your sum as $S_k=\sum_{j=1}^ka_j$ where, as you mention, $$\frac{a_{j+1}}{a_j} =\frac14\biggl(\frac{k-1}{j-1}-1\biggr)\biggl(\frac{n-k-1}{j-1}-1\biggr)$$ monotonically decreases (as the function of $j$) from $\infty$ to $0$. This means that $a_j$ first increases for $j\le m$ and then decreases for $j\ge m$, where $m=m(k,n)$ is determined by $a_{m+1}/a_m\approx 1$. In the your example, $m-1$ is a solution of the quadratic equation. Then the main term of the asymptotics is fully determined by the growth of $a_m$ in the sense $$\lim_{k\to\infty}S_k^{1/k} =\lim_{k\to\infty}a_{m(k,n)}^{1/k},$$ and then you can use Stirling's formula to determine the latter limit explicitly. (de Bruijn's book also contains details on how to compute more accurate asymptotics.)
Alternatively, you can use Zeilberger's algorithm of creative telescoping to find a recurrence equation for $S_k$; then the same asymptotics can be read from its characteristic equation.