Q:

Limit $\lim_{n→∞}\frac1n\sum_{j=c_n}^n\frac1{2n-2j+1}\sum_{k=1}^jk\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}.  $

If $c_n/n\to c>0$ as $n\to\infty$ (At the beginning of this question, I though the assumption is $c_n\to c$. With the help of metamorphy, the question will be ill-posed.  So, the assumption has changed to the current one.),  I would like to find$$\lim_{n→∞}\frac1n\sum_{j=c_n}^n\frac1{2n-2j+1}\sum_{k=1}^jk\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}. 
$$
Let $A_n=\sum_{j=c_n}^n\frac1{2n-2j+1}\sum_{k=1}^jk\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}$. Noting that 
$$
\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}=\frac{2n-2j+1}{2n-2k+1}\frac{(-1)^{n-j}{-1/2\choose n-j}}{(-1)^{n-k}{-1/2\choose n-k}}, 
$$
we have
\begin{align*}
&A_n\\
=&\sum_{j=c_n}^{n}\sum_{k=1}^{j}\frac{k(-1)^{n-j}{-1/2\choose n-j}}{(2n-2k+1)(-1)^{n-k}{-1/2\choose n-k}}\\
=&\sum_{k=1}^{c_n-1}\sum_{j=c_n}^{n}\frac{k(-1)^{n-j}{-1/2\choose n-j}}{(2n-2k+1)(-1)^{n-k}{-1/2\choose n-k}}+\sum_{k=c_n}^{n}\sum_{j=k}^{n}\frac{k(-1)^{n-j}{-1/2\choose n-j}}{(2n-2k+1)(-1)^{n-k}{-1/2\choose n-k}}\\
=&\sum_{k=1}^{c_n-1}\frac{k}{(2n-2k+1)(-1)^{n-k}{-1/2\choose n-k}}\sum_{j=c_n}^{n}(-1)^{n-j}{-1/2\choose n-j}+\sum_{k=c_n}^{n}\frac{k}{(2n-2k+1)(-1)^{n-k}{-1/2\choose n-k}}\sum_{j=k}^{n}(-1)^{n-j}{-1/2\choose n-j}\\
=&\sum_{k=n-c_n+1}^{n-1}\frac{n-k}{(2k+1)(-1)^{k}{-1/2\choose k}}\sum_{j=0}^{n-c_n}(-1)^{j}{-1/2\choose j}+\sum_{k=0}^{n-c_n}\frac{n-k}{(2k+1)(-1)^{k}{-1/2\choose k}}\sum_{j=0}^{k}(-1)^{j}{-1/2\choose j}\\
=&\sum_{k=n-c_n+1}^{n-1}\frac{n-k}{(2k+1)(-1)^{k}{-1/2\choose k}}(2n-2c_n+1)(-1)^{n-c_n}{-1/2\choose n-c_n}+\sum_{k=0}^{n-c_n}\frac{n-k}{(2k+1)(-1)^{k}{-1/2\choose k}}(2k+1)(-1)^{k}{-1/2\choose k}\\
=&(-1)^{n-c_n}(2n-2c_n+1){-1/2\choose n-c_n}\sum_{k=n-c_n+1}^{n-1}\frac{n-k}{(2k+1)(-1)^{k}{-1/2\choose k}}+\frac{n(n+1)-c_n(c_n-1)}{2},
\end{align*}
where the third-to-last equality follows since $\sum_{u=0}^{a}(-1)^{u}{-1/2\choose u}=(-1)^{a}(2a+1){-1/2\choose a}$.
Is it possible to give a further closed form of $A_n$?  And find the limit of $A_n/n$ (expresses it in terms of $c$)?  Please give me some help.  Thanks you very much. 

A:

Since $\sum_{0<j<c_n}[\ldots]$ is bounded w.r.t.  $n$, the limit doesn't change if $c_n$ is replaced by $1$. Let
$$A_n=\sum_{j=1}^{n}\frac{1}{2n-2j+1}\sum_{k=1}^{j}k\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1};$$
using the "double factorial" notation
$$(2n)!!=\prod_{k=1}^{n}(2k),\quad(2n+1)!!=\prod_{k=1}^{n}(2k+1),\quad 0!!=1!!=(-1)!!=1,$$
we have $\prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}=\frac{(2n-2k)!!}{(2n-2j)!!}\frac{(2n-2j+1)!!}{(2n-2k+1)!!}$ hence
\begin{align*}
A_n&=\sum_{j=1}^{n}\frac{(2n-2j-1)!!}{(2n-2j)!!}\sum_{k=1}^{j}k\frac{(2n-2k)!!}{(2n-2k+1)!!} 
\\\color{gray}{[\text{switching summations}]}\quad&=\sum_{k=1}^{n}k\frac{(2n-2k)!!}{(2n-2k+1)!!}\sum_{j=k}^{n}\frac{(2n-2j-1)!!}{(2n-2j)!!} 
\\\color{gray}{[\text{replacing }j\text{ with }n-j]}\quad&=\sum_{k=1}^{n}k\frac{(2n-2k)!!}{(2n-2k+1)!!}\sum_{j=0}^{n-k}\frac{(2j-1)!!}{(2j)!!} 
\\\color{gray}{[\text{replacing }k\text{ with }n-k]}\quad&=\sum_{k=0}^{n-1}(n-k)\underbrace{\frac{(2k)!!}{(2k+1)!!}\sum_{j=0}^{k}\frac{(2j-1)!!}{(2j)!!}}_{\text{now we prove this is }=1}. 
\end{align*}
Using the "generating functions" approach and the binomial series,
$$\sum_{k=0}^{\infty}z^k\sum_{j=0}^{k}a_j=\frac{1}{1-z}\sum_{k=0}^{\infty}a_k z^k,\qquad\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}z^k=(1-z)^{-1/2},$$
we obtain (using the binomial series once more)
$$\sum_{k=0}^{\infty}z^k\sum_{j=0}^{k}\frac{(2j-1)!!}{(2j)!!}=(1-z)^{-3/2}=\sum_{k=0}^{\infty}\frac{(2k+1)!!}{(2k)!!}z^k$$
and finally $A_n=\sum_{k=0}^{n-1}(n-k)=n\color{blue}{(n+1)/2}$. [Therefore, the given limit is $\infty$.]

