Smallest constant in exponent so that limit of sum is $0$ I am trying to work out the smallest constant $c>0$ so that $$\lim_{n \to \infty} \sum_{a=1}^n \sum_{b=0}^n {n \choose a} {n-a \choose b} \left({a+b \choose a} 2^{-a-b}\right)^{c n/\ln{n}} =0 .$$
If $c =3$ I can see that it certainly does and if $c=2$ it appears not to. How can you work out the constant exactly?
My current guess from a crude use of Stirling's approximation is that $c> 6/e$ is necessary and sufficient. 
 A: Let us first give the double summation a name: 
$$S_n = \sum_{a=1}^n \sum_{b=0}^n {n \choose a} {n-a \choose b} \left({a+b \choose a} 2^{-a-b}\right)^{c n/\ln{n}}.$$
Now looking at the binomial coefficient ${n-a \choose b}$ we see that the terms with $b > n - a$ play no role. So we could also have taken the sum from $b = 0$ to $n - a$:
$$S_n = \sum_{a=1}^n \sum_{b=0}^{n-a} {n \choose a} {n-a \choose b} \left({a+b \choose a} 2^{-a-b}\right)^{c n/\ln{n}}.$$
Next, we can see more structure if we change the indices of the double summation from summing over $a,b$ to summing over $k = a+b$ and $a$. Since $k$ runs from $1$ to $n$ and $a$ runs from $1$ to $k$ we then get
$$S_n = \sum_{k=1}^n \sum_{a=1}^k {n \choose a} {n-a \choose k-a} \left({k \choose a} 2^{-k}\right)^{c n/\ln{n}}.$$
We can further simplify the initial two binomials, by noting that $\binom{n}{a} \binom{n-a}{k-a} = \frac{n!}{a!(k-a)!(n-k)!} = \binom{n}{k} \binom{k}{a}$:
$$S_n = \sum_{k=1}^n {n \choose k} \sum_{a=1}^k {k \choose a} \left[{k \choose a} 2^{-k}\right]^{c n/\ln{n}}.$$
Considering the term inside the square brackets, we can get an upper bound on $S_n$ by replacing $a$ by $k/2$, and using $\binom{k}{k/2} \sim 2^k \cdot \sqrt{2 / (\pi k)}$ as follows:
$$\begin{align}
S_n &\leq \sum_{k=1}^n {n \choose k} \sum_{a=1}^k {k \choose a} \left[\max_A {k \choose A} 2^{-k}\right]^{c n/\ln{n}} \\
 &= \sum_{k=1}^n {n \choose k} \left[{k \choose k/2} 2^{-k}\right]^{c n/\ln{n}} \sum_{a=1}^k {k \choose a} \\
 &\sim \sum_{k=1}^n {n \choose k} 2^k \left[\sqrt{\frac{2}{\pi k}}\right]^{c n/\ln{n}} \\
 &= \sum_{k=1}^n {n \choose k} 2^k \exp\left(\frac{c n \ln(\frac{2}{\pi k})}{2 \ln n}\right) \\
 &= \exp\left(\frac{c n \ln(2/\pi)}{2 \ln n}\right) \sum_{k=1}^n {n \choose k} 2^k \exp\left(- \frac{c n \ln k}{2 \ln n}\right) = T_n \\
&\left[= \exp\left(\frac{c n \ln(2/\pi)}{2 \ln n}\right) \sum_{k=1}^n {n \choose k} \exp\left(k \ln 2 - \frac{c n \ln k}{2 \ln n}\right) \right]
\end{align}$$
Note in the last expression that the terms that will dominate the summation are those terms with $k \gg 0$; then both the exponential term and the binomial coefficients increase. On the other hand, when $k > n/2$ becomes too big, then at some point the terms will start to decrease again. But it is clear that the summation is dominated by terms with $k = O(n)$, and not the terms with $k = o(n)$. So if we replace the $\ln k$ by $\ln O(n)$ (but not the $2^k$ by $2^{O(n)}$), we can get an estimate for the upper bound $T_n$ as:
$$\begin{align}
S_n \leq T_n &\approx \exp\left(\frac{c n \ln(2/\pi)}{2 \ln n}\right) \sum_{k=1}^n {n \choose k} 2^k \exp\left(- \frac{c n \ln O(n)}{2 \ln n}\right) \\
 &\approx \exp\left(\frac{c n \ln(2/\pi)}{2 \ln n} - \frac{c n \ln O(n)}{2 \ln n}\right) \sum_{k=0}^n {n \choose k} 2^k \\
 &\stackrel{(a)}{=} \exp\left(-\frac{c n}{2} + O\left(\frac{n}{\ln n}\right)\right) 3^n \\
 &= \exp\left(\left(\ln(3)-\frac{c}{2}\right) n + O\left(\frac{n}{\ln n}\right)\right) \\
\end{align}.$$
Note that in (a), I used the fact that if $k = c_1 n$ for $0 < c_1 < 1$, then $\ln k = \ln c_1 n = \ln c_1 + \ln n = \ln n + O(1)$, meaning that this extra term disappears in the order term $O(n / \ln n)$. Finally, we want that $S_n \leq T_n = o(1)$, which means that $(\ln 3 - \frac{c}{2}) n = o(1)$, or in terms of $c$: $$c > \ln 9 \approx 2.19722.$$ 
Since $S_n \leq T_n$ and $T_n$ goes to zero for $c > \ln 9$, this shows that for $S_n = o(1)$ it suffices to take $c > \ln 9$. If $T_n$ is asymptotically tight then this bound is also necessary.

To verify the above result, the plot below shows numerics for $S_n$ for $\color{red}{c = 2.19}$ and $\color{blue}{c = 2.20}$. For $c = 2.19$ indeed it seems that $S_n \not\to 0$ for large $n$, while for $c = 2.20$ it looks like the curve tends to $0$ as $n \to \infty$. This does not prove anything, but since the red curve goes up at the end it does seem very likely that the turning point is above $2.19$, and since the blue curve seems to "accelerate" towards $0$ at $n \approx 200$, the turning point is probably below $2.20$ as well.

