# How to calculate this limit (if exist)?

I have just asked the calculation of the following summation see here $$S(a,b,m,n_1,n_2)=\sum_{k=0}^m a^k b^{m-k} {n_1\choose k} {n_2\choose m-k},$$ which is motivated by the calculation of the following limit (if exist) $$L_0=\lim_{m \to \infty}\frac{\sqrt{ab}S(a,b,m-1,n_1-1,n_2-1)}{S(a,b,m,n_1,n_2)}$$ where $$a>0, b>0, \frac{m}{n_1},\frac{m}{n_2}$$ are kept fixed.

My question is: Whether the above limit exists ($$a\neq b$$)? If it exists, then how to calculate it? When $$a=b$$, the result is simple $$\lim_{m \to \infty}\frac{a\cdot a^{m-1} {n_1+n_2-2\choose m-1}}{a^{m} {n_1+n_2\choose m}}=\lim_{m \to \infty}\frac{m(n_1+n_2-m)}{(n_1+n_2)(n_1+n_2-1)}$$ and this limit obviously exists, where we have used $$S(a,b,m,n_1,n_2)=a^m {n_1+n_2\choose m}$$ for $$a=b$$.

Despite the nice comments and answers in the previous question, I still don't know how to calculate the limit.

• Comment 1: $\lim_{c\to1}L_0=\frac{\nu_1+\nu_2-1}{(\nu_1+\nu_2)^2}$, where we have used $\lim_{c\to1}\kappa=\frac{\nu_1}{\nu_1+\nu_2}$ (see Pinelis's answer). This is consistent with the limit for $a=b$ in the above question: $\lim_{c\to1}L_0(c, \nu_1, \nu_2)=L_0(1, \nu_1, \nu_2)$, meaning that $L_0=L_0(c, \nu_1, \nu_2)$ is continuous at $c=1$.
– Dian
Oct 1 at 19:02
• Comment 2: $\lim_{c\to0}L(c, \nu_1, \nu_2)=L(0, \nu_1, \nu_2)=\frac{1}{\nu_2}$, where $\lim_{c\to0}\kappa=0$ (see the definition of $L$ in Pinelis's answer, and we have $L_0=\sqrt{c}L$). Note that $c=0$ means $a=0, b\neq0$.
– Dian
Oct 1 at 20:34
• Comment 3: $\lim_{c\to\infty}L_0(c, \nu_1, \nu_2)=L_0(\infty, \nu_1, \nu_2)=0$: When $c\to\infty$, we have $1-\kappa\sim\sqrt{\nu_2/c}$ for $\nu_1=1$ and $1-\kappa\sim\nu_2/(c(\nu_1-1))$ for $\nu_1>1$, so $(1-\kappa)(\nu_1-\kappa)\sim\nu_2/c$, and finally $L_0\sim1/(\nu_1\sqrt{c})$. Note that $c=\infty$ means $a\neq0, b=0$.
– Dian
Oct 1 at 22:11
• Remark: The summation has the symmetry $S(a,b,m,n_1,n_2)=S(b,a,m,n_2,n_1)$, which leads to the symmetry property of the limit: $L_0(c,\nu_1,\nu_2)=L_0(1/c,\nu_2,\nu_1)$ . And from the above 3 comments, $L_0$ at $c=1,0, \infty$ indeed satisfy this symmetry property. Especially, $L_0\sim\sqrt{c}/\nu_2 (c\to0)$ is equivalent to $L_0\sim1/(\nu_1\sqrt{c}) (c\to\infty)$ because of the symmetry property.
– Dian
Oct 2 at 18:55
• From the above arguments, $L_0$ as a function of $c$ is continuous for $c\in[0,\infty]$, where $c=\infty$ means $a\neq0, b=0$. Note that $L_0=0$ at $c=0, \infty$ and $L_0>0$ for $c\in(0,\infty)$, so $L_0$ has a maximum value at some $c$, for fixed $\nu_1, \nu_2$, and one can indeed check this by using the explicit expression of $L_0$. Especially, when $\nu_1=\nu_2$ is fixed, $L_0$ achieves the maximum at $c=1$, which is consistent with the symmetry property: $L_0(c)=L_0(1/c)$, so $L'_0(c)=-L'_0(1/c)/c^2$, so $L'_0(1)=-L'_0(1)$, and finally $L'_0(1)=0$.
– Dian
Oct 10 at 19:27

$$\newcommand\ka\kappa\newcommand{\de}{\delta}\newcommand\ep\varepsilon$$The problem obviously reduces to this one: find $$$$L:=\lim_{m\to\infty}\frac{S(c,m,n_1,n_2)}{S(c,m-1,n_1-1,n_2-1)},$$$$ where $$c:=a/b\ne1$$ and $$$$S(c,m,n_1,n_2):=\sum_{k=0}^m c^k A_k(m,n_1,n_2),$$$$ where $$$$A_k(m,n_1,n_2):=\binom{n_1}k \binom{n_2}{m-k}.$$$$

In accordance with the linked post, assume that $$m and $$m, so that $$m\to\infty$$, $$$$n_1/m\to\nu_1\in[1,\infty),\quad n_2/m\to\nu_2\in[1,\infty).$$$$

By considering the ratios $$r_k:=c^{k+1}A_{k+1}(m,n_1,n_2)/(c^k A_k(m,n_1,n_2))$$, one sees that $$$$S(c,m,n_1,n_2)\sim\sum_{k\sim \ka m} c^k A_k(m,n_1,n_2),$$$$ where $$$$\ka:=\frac{c \nu _1-\sqrt{\left(c \nu _1+c+\nu _2-1\right){}^2-4 (c-1) c \nu _1}+c+\nu _2-1}{2 (c-1)}\in(0,1).$$$$ (More specifically, note that $$r_k\ge1$$ for $$k\le k_*$$ and $$r_k\le1$$ for $$k\ge k_*$$, for a certain integer $$k_*\sim\ka m$$. Next, note that for each real $$\ep>0$$ there is some $$\de>0$$ such that for all large enough $$m$$ and all $$k=0,\dots,m-1$$ we have the implication if $$k/m-\ka\ge\de\implies r_k<1-\ep$$. So, for integers $$k\ge(\ka+3\de)m$$ we will have $$c^k A_k(m,n_1,n_2)\le(1-\ep)^{\de m}c^{k_*+1} A_{k_*+1}(m,n_1,n_2)$$ and hence
$$\sum_{k\ge (\ka+3\de)m} c^k A_k(m,n_1,n_2)\le m (1-\ep)^{\de m} c^{k_*+1} A_{k_*+1}(m,n_1,n_2) =o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$$.

Similarly, $$\sum_{k\le (\ka-3\de)m} c^k A_k(m,n_1,n_2) =o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$$.

Finally here, note that we can choose $$\de>0$$ to be however small.)

Also, for $$k\sim \ka m$$, $$$$\frac{A_k(m-1,n_1-1,n_2-1)}{A_k(m,n_1,n_2)} =\frac{(m-k)(n_1-k)}{n_1 n_2}\sim\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}.$$$$ So, $$$$L=\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}.$$$$

• A further remark: Based on Prof.Pinelis's excellent result, the original limit in my question is $L_0=\sqrt{c}\frac{(1-\kappa )(\nu_1-\kappa)}{\nu_1\nu_2}$. From the original definition, we see that the limit $L_0$ should be invariant under transformation $c\to 1/c, \nu_1\to\nu_2, \nu_2\to\nu_1$, and one can indeed check this. Note that $\kappa=\kappa(c, \nu_1,\nu_2)$ is a function of $c, \nu_1, \nu_2$.
– Dian
Oct 1 at 17:31
• Hi, I have one more question: Why $S(c, m, n_1, n_2)$ can be approximated by the terms $\sum_ {k\sim \kappa m} c^k A_k(m,n_1,n_2)$ ? In other words, from your answer, I know that the terms $c^k A_k(m,n_1,n_2)$ for $k\sim \kappa m$ are the largest terms in the summation, but why the rest smaller terms like $\sum_ {k\nsim \kappa m} c^k A_k(m,n_1,n_2)$ can be dropped? Thank you.
– Dian
Oct 5 at 22:35
• @Dian : I have added details on this. Oct 6 at 2:41