Limit of a sum with binomial coefficients Let $$A_k = \frac{\sum_{i=1}^ki{2k-i-1 \choose i-1}{i-1 \choose k-i}}{k{2k-1\choose k}}$$
$$B_k = \frac{\sum_{i=1}^ki{2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$
$$C_k = \frac{\sum_{i=1}^k(2k-2i-1){2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$
for $k\in\mathbb{N}$, where the binomial coefficients are to be taken as zero if any of the parameters are negative.
I want to prove that $S_k:=A_k+B_k+C_k$ is decreasing from $k=3$ and $S_k\to2/3$ as $k\to\infty$. I have been struggling with a formal mathematical proof for a few days, and I hope that somebody can point me to the right direction.
Note that based on the first 10000 values, the above statements seem to hold, and $A_k,B_k$ and $C_k$ seem to tend to $2/9$ as $k\to\infty$, furthermore, $A_k$ and $B_k$ are decreasing whereas $C_k$ is increasing from $k=3$. Also note that $B_k+C_k$ is simply
$$\frac{\sum_{i=1}^k(2k-i-1){2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}.$$
The reason for not making this simplification is that I found it interesting that each of $A_k,B_k$ and $C_k$ tends to $2/9$. It may be better to handle $B_k+C_k$ as a unite.
Motivation: This question is related to a preceding question. In the setting explained in the other question, $S_k$ is the probability of the marked red ball staying red in a random permutation of $(2k-1)$ balls. The above statements are already proven in an excellent answer to the preceding question. The aim of the present question is to give a new proof using $A_k,B_k,C_k$.
 A: Simplifying we have:
$$S_k = \frac{\sum_{i=1}^k i\binom{k}{2i-k}\binom{2k - i - 1}{k - 1}}{k\binom{2k-1}{k}}.$$
It follows that $S_k$ equals the coefficient of $x^k$ in
$$\frac{1+2x}{2(1+x)(1-x)^k\binom{2k-1}{k}}.$$
Using Lagrange inversion, we further have it as the coefficient of $x^k$ in $\frac{f(x)}{\binom{2k-1}{k}}$, where
$$f(x):=\frac{1+x+(1+2x)(1-4x)^{-1/2}}{2(2+x)}.$$
For $k\geq 1$, we further have
$$\frac1{k\binom{2k-1}{k}} = \int_0^1 t^{k-1}(1-t)^{k-1} dt$$
and thus
$$S_k = [x^{k-1}]\int_0^1 f'(xt(1-t))dt.$$
Computing this integral, we get the generating function:
$${\cal S}(x):=\sum_{k\geq 1} S_k x^k = \frac{x^{1/2}}{(8+x)^{3/2}}\left(4\operatorname{arctanh}\frac{x^{1/2}}{(8+x)^{1/2}} - 2\operatorname{arctanh}\frac{x-1-x^{1/2}(8+x)^{1/2}}3 + 2\operatorname{arctanh}\frac{x-1+x^{1/2}(8+x)^{1/2}}3\right) + \frac{2}{3(1-x)} - \frac{16}{3(8+x)}.$$
This function has poles at $x=1$ and $x=-8$ and thus $$\lim_{k\to\infty} S_k = \lim_{x\to 1}(1-x){\cal S}(x) = \frac23.$$
A: It appears that
\begin{equation*}
    S_k=\frac23+\frac ak+\frac b{k^{3/2}}+\frac c{k^2}+\frac{d_k}{k^{5/2}},\tag{1}\label{1}
\end{equation*}
where $a,b,c$ are certain real numbers such that $a>0$ and the $|d_k|$'s are bounded by a certain real $d$.
A proof of \eqref{1} should be rather straightforward (even if quite tedious) using Stirling's formula and the Laplace method -- both with higher-order terms but with explicit bounds on the remainder terms everywhere, as well as the Euler--Maclaurin summation formula.
The first steps in this direction can be the observations that (i)
\begin{equation*}
    S_k=\sum_{i=1}^k a_{k,i} 
\end{equation*}
with
\begin{equation*}
    a_{k,i}:=\frac{i (2 k-i-1)!}{(2 i-k)! (k-i)! (2 k-2 i)!}\Big/ \binom{2 k-1}{k}
\end{equation*}
and (ii) $a_{k,i+1}\ge a_{k,i}$ if $i\le\frac23\,k-1$ and $a_{k,i+1}\le a_{k,i}$ if $i\ge\frac23\,k$.
It will then follow from \eqref{1} that $S_k$ is decreasing in $k\ge k_{a,b,c,d}$, where $k_{a,b,c,d}$ depends only on $a,b,c,d$. If $k_{a,b,c,d}$ is not too large, it should then be easy to check that $S_k$ is decreasing in $k$ if $3\le k\le k_{a,b,c,d}$. Thus, you will get that $S_k$ is decreasing in all $k\ge3$, to $2/3$, as desired.
A: Jorge Zuniga showed the identity
\begin{equation*}
    S_{n+2}=p_n\,S_{n+1}+q_n\,S_n, \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
    p_n=\frac{21 n^2 + 44 n + 16}{24 n^2 + 52 n + 24},\quad q_n=\frac{3 n^2 + 8 n + 5}{24 n^2 + 52 n + 24}. 
\end{equation*}
Based on \eqref{1}, Jorge Zuniga showed that
\begin{equation*}
S_{n+1}<S_n \tag{2}\label{2}
\end{equation*}
for all large enough $n$.
Let us show that \eqref{2} holds for all $n\ge3$. Rewrite \eqref{1} as
\begin{equation*}
    R_{n+1}=p_n+q_n/R_n, \tag{3}\label{3}
\end{equation*}
where
\begin{equation*}
    R_n:=S_{n+1}/S_n. 
\end{equation*}
Note that
\begin{equation*}
    R_{1,n}<R_n<R_{5,n} \tag{4}\label{4}
\end{equation*}
for $n=10$, where
\begin{equation*}
    R_{a,n}:=1-\frac{1}{9 n^2+a n}. 
\end{equation*}
Let us prove by induction on $n$ that \eqref{4} holds for all $n\ge10$. Suppose \eqref{4} holds for some $n\ge10$.
For all $n\ge10$ we have
\begin{equation*}
    R_{1,n+1}\le p_n+q_n/R_{5,n},\quad p_n+q_n/R_{1,n}\le R_{5,n+1}
\end{equation*}
and hence, by \eqref{3} and \eqref{4},
\begin{equation*}
    R_{1,n+1}\le p_n+q_n/R_{5,n}<R_{n+1}<p_n+q_n/R_{1,n}\le R_{5,n+1},
\end{equation*}
so that \eqref{4} holds with $n+1$ in place of $n$.
So, for all $n\ge10$ inequalities \eqref{4} hold and hence $R_n<R_{5,n}<1$, which implies \eqref{2} -- for $n\ge10$. It is easy to see that \eqref{2} holds for $n=3,\dots,9$ as well.
Thus, \eqref{2} holds for all $n\ge3$, as desired.
