Maximize a Function of Binomial PMF's with respect to $n$ Let 
    $$
    f(k,n,p) = {n \choose k}p^k(1-p)^{n-k}
$$
be the binomial probability mass function. I want to maximize a function of binomial pmf's with respect to $n$:
    \begin{align*}
    g(n) \equiv &\left(1-\sum_{T=0}^{L}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B) \right)\\ \cdot &\left( \sum_{T=0}^{U}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B) \right)
\end{align*}
for fixed $L,U,N,\in \mathbb{Z}^+,\ p_A,p_B \in [0,1]$. In the case of a single binomial pmf, $f(n,k,p)$ is unimodal with respect to $n$ so one can find $n^* \equiv \operatorname{argmax} f(k,n,p)$ by finding $n$ such that:
    $$
    f(k,n,p) \geq f(k,n+1,p)\qquad \text{and}\qquad f(k,n,p)\geq f(k,n-1,p)
$$
which yields
    $$
    \frac{k}{p} \geq n\geq \frac{k}{p}-1;
$$
i.e., $f(k,n,p)$ for fixed $k,p$ is maximized by
    $$
   n^* =  \text{floor}\left(\frac{k}{p}\right).
$$
Having tested empirically, I think $g(n)$ is also unimodal with respect to $n$, but haven't been able to prove it. Additionally, I have not been able to solve the analogous inequalities:
    $$
    g(n) \geq g(n+1)\qquad g(n)\geq g(n-1)
$$
(neither by hand nor using Mathematica).
Any suggestions for how to maximize $g(n)?$ I'd prefer to avoid differentiating if possible: I think a solution which only deals with integer values is more likely to provide illuminating intuition for the problem I'm working on, and gamma functions are well-defined on values that aren't admitted by my problem. That being said, if it's the only way to do this, so be it.
Edit for clarification: I am primarily interested in the case $L < U \leq N$.
 A: Partial Answer
Let $m=\min(L,U),\;M=\max(L,U),$ and 
$$h(n)=\sum_{T=0}^{m}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B). $$
 We have three cases:
Case 0: $L=m=U.$ Then
$$g(n)=(1-h(n))\,h(n), $$
which is minimized by $h(n)=1/2,$ if this is possible. 
Case 1: $L=m<U.$ Then
\begin{align*}
    g(n) = \left(1-h(n) \right) \cdot \left( h(n)
+\sum_{T=m+1}^{M}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B) \right).
\end{align*}
This is analogous to shifting the roots of the function $f(x)=x(1-x)$ as $\tilde{f}(x)=(x+a)(1-x);$ we can still maximize this fairly easily by noting that we want $x$ to be half-way in-between the two roots $1$ and $-a:$ $x=(1+(-a))/2=(1-a)/2.$ Therefore, we want our solution to be
$$h(n)=\frac{1-\sum_{T=m+1}^{M}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B)}{2}.$$
Case 2: $L>m=U.$ Then we have
$$ g(n)=\left(1-h(n)-\sum_{T=m+1}^{M}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B)\right)h(n).$$
This is analogous to shifting our root like this: $\hat{f}(x)=x(1-x-a),$ with the maximum at the average of the two roots $0$ and $1-a,$ yielding $(1-a)/2,$ as before. So, our maximum would occur at
$$h(n)=\frac{1-\sum_{T=m+1}^{M}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B)}{2}, $$
exactly as before.
To sum up: the minimum occurs at
$$h(n)=\begin{cases}1/2,\quad& L=U\\ \left(1-\sum_{T=m+1}^{M}\sum_{t=0}^T f(t,n,p_A) \cdot f(T-t,N-n,p_B)\right)/2,\quad &L\not=U\end{cases}. $$
What my answer does not do is show you how to solve these $h(n)=$ -type equations. You might have to do that numerically; you might also not get an exact answer, given the integer nature of these expressions (at least with respect to the sum limits).
