Limit of ratio of a combinatorial expression? The following function takes as arguments $n$ and
$\left\{n_{i}\right\} = \left\{n_{0}, n_{1}, n_{2}\right\}$:
$$
\operatorname{f}\left(n,\left\{n_i\right\}\right) =
\sum_{j = 0}^{n}\binom{n}{j}n_{0}^{\,\,\left(n - j\right)\left(n - j - 1\right)/2}\,\,\,\,\,\,
n_{1}^{\left(n - j\right)\,j}\,\,
n_{2}^{\,\,j\,\left(j - 1\right)/2} 
$$
Is there any hope of calculating the following limit
$$
\lim_{n \to \infty}\frac{\operatorname{f}\left(n,\left\{n_{i}\right\}\right)}
{\operatorname{f}\left(n,\left\{m_{i}\right\}\right)}
$$ in terms of $n_{0}, n_{1}, n_{2}$ and $m_{0},m_{1},m_{2}\ ?$.

The limit needs to be calculated for a given
$n_{0}, n_{1}, n_{2}$ and $m_{0},m_{1},m_{2}$ i.e. they are constant.
 A: Notice $(n-j)(n-j-1)/2+(n-j)j+j(j-1)/2=n(n-1)/2$, let $a=\frac{n_1}{n_0}, b=(\frac{n_2}{n_0})^{\frac{1}{2}}$, then
\begin{equation}
f=n_0^{n(n-1)/2} \sum_j \binom{n}{j} (a^{n-j}b^{j-1})^j.
\end{equation}
Let $\alpha = \text{min } \{a, b\}$, $\beta = \text{max } \{a, b\}$, then $\alpha^{n-1} \le a^{n-j}b^{j-1} \le \beta^{n-1}$,
\begin{equation}
n_0^{n(n-1)/2}(1+ \alpha^{n-1})^n \le f \le n_0^{n(n-1)/2}(1+ \beta^{n-1})^n.
\end{equation}
(Special cases: $\alpha=\beta$ or $\beta<1$. If $\beta < 1$, $f$ will be very close to $n_0^{n(n-1)/2}$.)
We can just take one term of that summation, in some cases it will give a better lower bound than above.

*

*take $j=n$ gives $f>n_0^{n(n-1)/2} b^{n(n-1)}$, if $b>1$ and $b>a$ this shows $f$ is relatively close to $n_0^{n(n-1)/2} b^{n(n-1)}$.


*take $j=n/2$ gives $f>n_0^{n(n-1)/2} a^{\frac{1}{4}n^2} b^{\frac{1}{4}n(n-2)}$.
Just very small improvement, it is unlikely to solve for all $n,m$, but seems very likely that limit almost always exists and is $0$ or $\infty$ or $1$ for some special cases.
