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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.

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  • $\begingroup$ 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 $c=1$. $\endgroup$
    – Dian
    Oct 1 at 19:02
  • $\begingroup$ 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 $a=0, b\neq0$. $\endgroup$
    – Dian
    Oct 1 at 20:34
  • $\begingroup$ 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 $a\neq0, b=0$. $\endgroup$
    – Dian
    Oct 1 at 22:11
  • $\begingroup$ 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. $\endgroup$
    – Dian
    Oct 2 at 18:55
  • $\begingroup$ 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 $L'_0(1)=0$. $\endgroup$
    – Dian
    Oct 10 at 19:27

1 Answer 1

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

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

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 \begin{equation} S(c,m,n_1,n_2)\sim\sum_{k\sim \ka m} c^k A_k(m,n_1,n_2), \end{equation} where \begin{equation} \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). \end{equation} (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 $k\sim \ka m$, \begin{equation} \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}. \end{equation} So, \begin{equation} L=\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}. \end{equation}

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  • $\begingroup$ 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 $c, \nu_1, \nu_2$. $\endgroup$
    – Dian
    Oct 1 at 17:31
  • $\begingroup$ 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. $\endgroup$
    – Dian
    Oct 5 at 22:35
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    $\begingroup$ @Dian : I have added details on this. $\endgroup$ Oct 6 at 2:41

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