I am, through some combinatorial problems which I'm working on, trying to figure out what the following sum becomes as $n\rightarrow \infty$:
\begin{equation*} \sum_{i=1}^{n-1} \binom{n}{i}3^{\binom{n-i}{2}-\binom{n}{2}}2^{\binom{i}{2}} \end{equation*}
The main question is: Does it go to infinity, $0$ or something else? Can we at least exclude some of the cases (like it going to 0)?
First note that $${n-i \choose 2}-{n \choose 2} = \frac{(n-i)(n-i-1)-n(n-1)}{2}=\frac{-2in+i^2+i}{2}$$
Thus \begin{align} 3^{{n-i \choose 2}-{n \choose 2}}2^{{i \choose 2}} &= 3^{\frac{-2in+i^2+i}{2}}2^{\frac{i(i-1)}{2}} \\ &= 3^{-in} \cdot 3^{\frac{i^2+i}{2}} \cdot 2^{\frac{i(i-1)}{2}} \\ &< 3^{-in} \cdot 6^{\frac{i^2+i}{2}} \\ &< 3^{-in} \cdot 6^{\frac{in}{2}} \\ &= \left(0.5\right)^{-in} \\ &< n^{-i} \end{align}
Since $(n+\frac{1}{n})^n = \sum n^{-i} {n \choose i}$ goes to $e$, we have that the given sum is smaller than or equal to Euler's number.