Sum to infinity

$z^5-2z^4-2z^3-4z^2-5z$ $=(z^3-3z^2+z-5)(z+1)z$ Given expression simplifies to: $\displaystyle\sum_{z=1}^{\infty}\frac{1}{z(z+1)} \quad\quad \small\color{#20A900}{\text{(Telescopic Series)}}$ $\quad\quad\quad\quad \Large{=1}$

Problem:

$\displaystyle\sum_{z=1}^{\infty}\frac{z^3-3z^2+z-5}{z^5-2z^4-2z^3-4z^2-5z}=\,?$

Solution:

$\displaystyle\sum_{z=1}^{\infty}\frac{z^3-3z^2+z-5}{z^5-2z^4-2z^3-4z^2-5z}$

$= \, \displaystyle\sum_{z=1}^{\infty}\frac{z^3-3z^2+z-5}{z(z^4-2z^3-2z^2-4z-5)}$

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For $z=-1$ , we have:

$z^4-2z^3-2z^2-4z-5 \, = \, (-1)^4-2(-1)^3-2(-1)^2-4(-1)-5 \, = \, 1+2-2+4-5 \, = \, 0$ .

Therefore ( $z+1$ ) is a factor of $z^4-2z^3-2z^2-4z-5$ such that:

$z^4-2z^3-2z^2-4z-5=(z+1)(z^3-3z^2+z-5)$

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$= \, \displaystyle\sum_{z=1}^{\infty}\frac{z^3-3z^2+z-5}{z(z+1)(z^3-3z^2+z-5)}$

$= \, \displaystyle\sum_{z=1}^{\infty}\frac{1}{z(z+1)} \, \text{(A telescoping series of course)}$

$= \, \displaystyle\sum_{z=1}^{\infty}\frac{z+1-z}{z(z+1)}$

$= \, \displaystyle\sum_{z=1}^{\infty}\frac{z+1}{z(z+1)}-\frac{z}{z(z+1)}$

$= \, \displaystyle\sum_{z=1}^{\infty}\frac{1}{z}-\frac{1}{z+1}$

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Because the given telescoping series to $n$ terms can be given as: $\sum_{z=1}^n \frac{1}{z(z+1) } = \frac{1}{1} - \frac{ 1}{n+1}.$

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$= \, \frac{1}{1}-\displaystyle\lim_{z\rightarrow\infty}{\frac{1}{z+1}} \, = \, 01$

Conclusive Result:

$\displaystyle\sum_{z=1}^{\infty}\frac{z^3-3z^2+z-5}{z^5-2z^4-2z^3-4z^2-5z}=\,\boxed{01}$