Approximate $\sum\limits_{k=0}^{m-1}\frac{k}{m-k}$

Question

How do I approximate (or better find a closed formed formula) for $\sum\limits_{k=0}^{m-1}\frac{k}{m-k}$, where $m$ is large.

This looks suspiciously like a partial sum for the harmonic series, but I do know how to find a closed form for it. (Unless I missed something obvious).

Any help or insights is appreciated

Answer 1

$ \textbf{Hint :} $ \begin{aligned}\sum_{k=0}^{m-1}{\frac{k}{m-k}}&=\sum_{k=1}^{m}{\frac{m-k}{k}}\\ &=mH_{m}-m\end{aligned}

Answer 2

The closed-form solution is $$\gamma m-m+m \psi ^{(0)}(m)+1$$

using the Euler gamma and polygamma functions:

Answer 3

You have been given the exact result.

If you want a good approximation, using the asymptotics of harmonic numbers $$mH_{m}-m\sim m (\log (m)+\gamma -1)+\frac{1}{2}-\frac 1 {12m} \Bigg[1-\frac{55671 m^2+110200}{105 \left(5302 m^4+13020 m^2+3549\right)} \Bigg]$$ which is $O\left(\frac{1}{m^{11}}\right)$.

For $m=10$, the approximation is in an absolute error of $4.82\times 10^{-14}$