Finding a specific series sum $\sum_{k=1}^{i-1}(\frac{((ak^3+a(1-2i+2n)k^2+2b)}{2ak})^{-1}$

Question

I am stuck in this specific sum equation as below. Sum it $k$ is from $1$ to $i-1$ and others are static variables.

Please help to find $\sum_{k=1}^{i-1}(\frac{((ak^3+a(1-2i+2n)k^2+2b)}{2ak})^{-1}$

I appreciated your helps. Thank you.

Answer 1

You are considering $$S_i=\sum_{k=1}^{i-1} \frac{2ak}{ak^3+a(1-2i+2n)k^2+2b}$$ I think that what I should do is to write $$ak^3+a(1-2i+2n)k^2+2b=a(k-r)(k-s)(k-t)$$ Now using partial fraction decomposition $$ \frac{2ak}{ak^3+a(1-2i+2n)k^2+2b}$$ write $$\frac{2 r}{(r-s) (r-t) (k-r)}-\frac{2 s}{(r-s) (s-t) (k-s)}-\frac{2 t}{(r-t) (t-s) (k-t)}$$ to face three summations looking like $$I_x=\sum_{k=1}^{i-1} \frac 1 {k-x}=H_{i-x-1}-H_{-x}$$

The problem will be to find the expressions of the roots $(r,s,t)$ but from a formal point of view we have the result.