Finding the value of the sum $\sum_{k=1}^{\lceil\alpha\sqrt{n}\rceil}\frac{k}{2n-k}$ as $n\to\infty$ I am trying to find the exact value of
$$
\lim_{n\to\infty}\prod_{k=1}^{\lceil\alpha\sqrt{n}\rceil-1}\frac{2n-2k}{2n-k}
$$
for any $\alpha>0$. By some simple bounds I have found that
$$
e^{-\alpha^2}\lesssim\prod_{k=1}^{\lceil\alpha\sqrt{n}\rceil-1}\frac{2n-2k}{2n-k}\lesssim e^{-\alpha^2/4}\quad\text{as }n\to\infty,
$$
so I know that I am considering the correct asymptotics, i.e. there is asymptotically the correct number of factors in the product. My idea was then to show that I am not losing anything in my upper bound. Thus, I have also shown that
$$
\prod_{k=1}^{\lceil\alpha\sqrt{n}\rceil-1}\frac{2n-2k}{2n-k}\sim\exp\bigg(-\sum_{k=1}^{\lceil\alpha\sqrt{n}\rceil-1}\frac{k}{2n-k}\bigg),
$$
i.e. the two are asymptotically equivalent, meaning that their quotient tends to $1$ as $n\to\infty$. This leaves me with finding the limit of the sum. The $-1$ in the upper limit is asymptotically negligable, so we are looking to find
$$
\lim_{n\to\infty}\sum_{k=1}^{\lceil\alpha\sqrt{n}\rceil}\frac{k}{2n-k},
$$
if the limit exists, which I believe to be the case. I have tried to write it as a Riemann-sum as follows:
$$
\sum_{k=1}^{\lceil\alpha\sqrt{n}\rceil}\frac{\frac{k}{\sqrt{n}}}{2-\frac{k}{\sqrt{n}}\frac{1}{\sqrt{n}}}\cdot\frac{1}{\sqrt{n}},
$$
but alas, this is only almost a Riemann-sum, at the $k/\sqrt{n}$ in the denominator is divided by $\sqrt{n}$ too much.
WolframAlpha says, without explanation, that
$$
\sum_{k=1}^{m-1}\frac{k}{2n-k}=-2n\psi^{(0)}(m-2n)-n+2n\psi^{(0)}(1-2n)+1,
$$
where $m\leq 2n$ and $\psi^{(0)}$ is the digamma function. I have no clue how one would show this, but I'm guessing that this calculation would suffice in finding the limit.
Edit: found an error in the upper bound, which I have corrected.
 A: As for the sum $S_n=\sum_{k=1}^{\lceil\alpha\sqrt n\rceil}\frac k{2n-k}$.  For all $k$ in the range of summation, $0<k<1+\sqrt n \alpha$, so
$$ \frac k{2n}\le\frac k{2n-k}\le\frac k{2n-(1+\sqrt n \alpha)} = \frac k{2n}\times\frac{2n}{2n-(1+\sqrt n \alpha)}.$$
Hence $$ \frac1 {2n} \frac {\lceil\alpha\sqrt n\rceil(\lceil\alpha\sqrt n\rceil+1)}2 
\le S_n \le \frac1 {2n} \frac {\lceil\alpha\sqrt n\rceil(\lceil\alpha\sqrt n\rceil+1)}2
\times\frac{2n}{2n-(1+\sqrt n \alpha)}.$$
The upper and lower bounds both converge to $\alpha^2/4$.
A: @kimchilover's comment was correct, the approximation $k/(2n-k)\approx k/2n$ is sufficient. To quantify the error, we write
$$
\begin{align}
\frac{k}{2n-k}
&=\frac{k}{2n}+\left(\frac{k}{2n}\right)^2+\left(\frac{k}{2n}\right)^3+\dots
\\&=\frac{k}{2n}+O\left(\left(\frac{k}{2n}\right)^2\right)
\\&=\frac{k}{2n}+O\left(\frac{1}{n}\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^{\alpha \sqrt n}\frac{k}{2n-k}
&= \sum_{k=1}^{\alpha \sqrt n}\left(\frac{k}{2n}+O(1/n)\right)
\\&=\frac{(\alpha \sqrt n)^2}{2\cdot 2n}+O(1/\sqrt n)
\\&=\frac{\alpha^2}4+O(1/\sqrt n)
\end{align}
$$
Finally, we get
$$
\prod_{k=1}^{\lceil\alpha\sqrt{n}\rceil-1}\frac{2n-2k}{2n-k}
=\exp\bigg(-\frac{\alpha^2}4+O(1/\sqrt n)
\bigg)\sim \exp\left(\frac{-\alpha^2}{4}\right)
$$
