Given the following sum:
$\sum_{x=1}^n \frac{1}{bx - x^2}$
(b is an integer constant)
It appears to me that partial sum cannot be calculated without using Harmonic numbers, or Digamma function. From wolfram:
https://www.wolframalpha.com/input/?i=sum+1%2F(bx-x%5E2),+x%3D1+to+n
$\sum_{x=1}^n \frac{1}{bx - x^2} = \frac{-\psi^{(0)}(-b+n+1)+\psi^{(0)}(1-b)+\psi^{(0)}(n+1)+\gamma}{b} $
Is it possible that for some specific value of n, the partial sum would be calculable with a simpler formula. If not applicable to this sum, is it something that is possible for any other sum ? In my example, the partial sum i'm looking for is
$\sum_{x=1}^\frac{b}{2} \frac{1}{bx - x^2}$
You can see in the chart that everything is kind of symmetric, witch is why i was under the impression that the sum (orange line) up to x=1, x=b/2 and x=b-1 would be easier to calculate than the other points. We also know that
$\sum_{x=1}^\frac{b}{2} \frac{1}{bx - x^2} = \frac{1}{2}\sum_{x=1}^{b-1} \frac{1}{bx - x^2}+ \frac{2}{b^2}$
