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I'm studying Laurent Series, and I thought I understood it but after solving one exercises I get confused. The problem was to get the Laurent series for the following function:

[tex]\frac{1}{\left(z-1\right)^2\left(z+3\right)}[/tex]

I do it this way:

[tex]\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{z+3+1-1}=[/tex]

[tex]\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{4-\left(-\left(z-1\right)\right)}=[/tex]

[tex]\frac{1}{\left(z-1\right)^3}\cdot \frac{1}{1-\left(-\left(\frac{z-1}{4}\right)\right)}=[/tex]

[tex]\frac{1}{\left(z-1\right)^3}\cdot \sum_{n=0}^\infty \left(-\frac{4}{z-1}\right)^n=[/tex]

[tex]\sum_{n=0}^\infty \left(-4\right)^n\cdot \frac{1}{\left(z-1\right)^{n+3}}[/tex]

But when I saw the solution they say that is:

[tex]

\frac{1}{4\left(z-1\right)^2} - \frac{1}{4^2\left(z-1\right)}+\frac{1}{4^3}-\frac{\left(z-1\right)}{4^4}+\frac{\left(z-1\right)^2}{4^5}+\dots

[/tex]

which is different from what I get. However I noticed that if I change the limits of the sum to -inf,0 the expression is equal. Why do this happens or what do I did wrong?