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}. $

Question

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.

Answer 1

[outdated, should be unaccepted and deleted]

Answer 2

Now we assume that $c_n\to c$ as $n\to\infty$. I am not familiar with double factorial, but I tried to follow your approach (using the identify ${{2k\choose k}}/2^{2k}=(-1)^{k}{{-1/2\choose k}}$ implying that \begin{align*} \prod_{i=k}^{j-1}\frac{2n-2i}{2n-2i+1}&=\frac{2n-2j+1}{2n-2k+1}2^{2(j-k)}\frac{{2n-2j\choose n-j}}{{2n-2k\choose n-k}}\\ &=\frac{2n-2j+1}{2n-2k+1}\frac{(-1)^{n-j}{{-1/2\choose n-j}}}{(-1)^{n-k}{{-1/2\choose n-k}}} \end{align*} ) and I obtained that (by switching the summations and changing the variables) $A_n=\alpha_n+\beta_n$ where $$ \alpha_n=\frac{(-1)^{n-c_n}(2n-2c_n+1){{-1/2\choose n-c_n}}}{n}\sum_{k=n-c_n+1}^{n-1}\frac{n-k}{(2k+1)(-1)^{k}{{-1/2\choose k}}} $$ and $$ \beta_n=\frac{n(n+1)-c_n(c_n-1)}{2n}. $$

Is it possible to simplify $A_n$ further so as to find the limit?