# The upper bound of a series

I have the following series: $$F(l, m) = \frac{(-1)^m}{2^{2l+m}}\sum_{j=0}^l C_l^jC_{m+l}^j(-3)^j,$$ where $l$, $m \in \mathbb N_+$ are positive integers and $C_l^j$ is the binomial coefficient. The problem is to find the largest value of $F(l,m)$.

I tried a lot of calculations and guess that $F(1,1) = 5/8$ reaches the maximum point, but I have no idea about the proof. Could anyone give me some idea to show this?

It seems that for fixed $l$, $|F(l,m)|$ first increases to its maximum and keep decreasing after that. And similar thing happens when $m$ is fixed. If one can find a number $N$ large enough such that the upper bound holds for all $l$, $m > N$, then the other cases can be checked by programming.

• $C_l^j$ is the binomial coefficent? Commented Dec 1, 2016 at 12:12
• yes. I edited it. Commented Dec 1, 2016 at 12:16
• Wolfram Mathematica gives the following expression for the $F(l,m)$ $$F(l,m) = (-1)^m 2^{-2 l-m} \, _2F_1(-l,-l-m;1;-3)$$ Commented Dec 1, 2016 at 13:25
• @uranix yes, i also find that, and the problem becomes to estimate the hypergeometric function 2F1[-l,-l-m,1,-3]; do you know where i can find some useful information about such an upper bound? Commented Dec 1, 2016 at 13:29

The function can be written as $$F(l,m) = (-1)^m 2^{-2l-m} {}_2F_1(-l, -l-m; 1; -3)$$ where ${}_2F_1(a,b;c;z)$ is the Hypergeometric function. Applying Pfaff's transformation $${}_{2}F_{1}(a,b;c;z)=(1-z)^{-a}{}_{2}F_{1}\left(a,c-b;c;{\tfrac {z}{z-1}}\right)$$ we obtain $$F(l,m) = \left(-\frac{1}{2}\right)^m {}_2F_1\left(-l, 1+l +m, 1; \frac{3}{4}\right)$$ The ${}_2F_1\left(-l, 1+l +m, 1; z\right)$ is a Jacobi polynomial $${}_{2}F_{1}(-l,1+m +l;1;z)=P_{l}^{(0 ,m)}(1-2z),$$ thus $$F(l,m) = (-2)^{-m} P_l^{(0,m)}\left(-\frac{1}{2}\right) = (-1)^{l+m} 2^{-m} P_l^{(m,0)}\left(\frac{1}{2}\right).$$
It seems that for fixed $l$ the function $|F(l,m)|$ oscillates when $m \lesssim 2l$ and quickly decays after $m > 2l$. Also $$\max_{m \in \mathbb N^+} |F(l,m)| < \max_{1 < m < \infty} |F(l,m)|$$ (global maximum is attained just before the decay) is a monotonically decreasing function of $l$, so the maximum is really at $l = m = 1$.