Evaluating $\lim_{n\to\infty}\sum_{k=0}^{n/2-1} \frac{k}{(n-k)^2} $

Question

I am trying produce a closed form for $\lim_{n\to\infty}S_n$ where:

$$ S_n = \sum_{k=0}^{n/2-1} \frac{k}{(n-k)^2} $$

For example, if $n = 10$:

$$S_n = \sum_{k=0}^{4} \frac{k}{(10-k)^2} = \frac{0}{10^2}+\frac{1}{9^2}+\frac{2}{8^2}+\frac{3}{7^2}+\frac{4}{6^2} = 0.2159 $$

If $n = 1000$: $S_n = 0.3059$

If $n = 10000$: $S_n = 0.3068$

Though simulation, as $n$ increases, this series converges to $\approx 0.3069$.

However, I would like to know how to calculate this value without resorting to plugging in large $n$. Even an approximation of the summation as a function of $n$ would be fine.

However, my knowledge of sum calculation is limited to geometric series and harmonic sequences.

What I've tried:

Series Analysis

I tried to find some patterns in the sequence, and know that the first term is always $0$, and the last term is always $\frac{n/2 -1}{(n/2 + 1)^2}$, which approaches $0$ as $n$ increases, but I'm not sure how to generate a closed form solution for $S_n$.

I also thought of breaking the series up into two parts as shown:

$$ \sum_{k=0}^{n/2-1} k \cdot \frac{1}{(n-k)^2} $$

So there would be an arithmetic progression ($S_{n+1} = S_{n} + 1 $), with each term multiplied by $1/{(n-k)}^2$, but this still has lead me nowhere.

I have noticed that if the series was:

$$ \sum_{k=0}^{n/2-1} \frac{n}{(n-k)^2} $$

($k$ in numerator replaced with $n$), the sum converges to $1$. I did this because this series is strictly larger than the original, and I wanted to use the direct comparison test to formally prove convergence. Of course, I don't know how to prove that this series converges either, so this wasn't much help.

Integration

The last thing I tried was to plot the values of each term in the summation, and then plot them for different $n$. If I could get an equation for these curves as a function of $n$, I should be able to integrate them and get the summation value, right? The curves look to have the general form:

$$ f(k) = y \cdot (e^{k\cdot x} - 1) $$

where $x$ and $y$ are two real numbers. Does this idea hold merit? How could I find these curves analytically?

MATLAB code used to calculate sum:

n = 1000;
sn = 0;
for k = 0:n/2-1
    sn(k+1) = k / (n-k)^2;
end
sum(sn)

Answer 1

We have the Riemann sum

$$\begin{align} \sum_{k=0}^{n/2-1}\frac{k}{(n-k)^2}&=\frac1n\sum_{k=0}^{n/2-1}\frac{k/n}{(1-k/n)^2} \end{align}$$

Hence,

$$\lim_{n\to\infty}\sum_{k=0}^{n/2-1}\frac{k}{(n-k)^2}= \int_0^{1/2}\frac{x}{(1-x)^2}\,dx=1-\log(2)$$