Partial sum formula of a polynomial series?

I am trying to find the partial sum formula of the following series:

$$\sum_{y=1}^{\infty} \frac{4y^2-12y+9}{(y+3)(y+2)(y+1)y}$$

I have tried using Faulhaber's formula without success. I have also tried rewriting the system using partial fraction decomposition to obtain 4 terms. This didn't solve the issue either.

When I use WolframAlpha to evaluate the sum, (or other computational software), it becomes 1/2. Is there some way to derive the this infinite sum to a partial sum formula?

This partial sum formula is according to WolframAlpha:

$$\frac{n^3-2n^2+3n}{2(n+1)(n+2)(n+3)}$$

Thank you in advance!
J

Answer

Decompose the fraction on simple elements:

$$\frac{4y^2-12y+9}{(y+3)(y+2)(y+1)y}=\frac{a}{y}+\frac{b}{y+1}+\frac{c}{y+2}+\frac{d}{y+3}$$

and since the series is convergent then we have $a+b+c+d=0$. Now the partial sum is

$$\sum_{y=1}^n \frac{a}{y}+\frac{b}{y+1}+\frac{c}{y+2}+\frac{d}{y+3}=a\sum_{y=1}^n \frac{1}{y}+b\sum_{y=2}^{n+2} \frac{1}{y}+c\sum_{y=3}^{n+3} \frac{1}{y}+d\sum_{y=4}^{n+4} \frac{1}{y}$$
and the simplification is clear. Can you take it from here?