<s>[INST] Taylor Series of $\frac{1}{3z+1}$ at $a=-2$ 
Find the Taylor Series of $\frac{1}{3z+1}$ at $a=-2$

When we are asked to find the taylor series at a point $p$ the taylor series should be at the form $\sum a_n(z-p)^n$?
So 
$$\frac{1}{3z+1}=\frac{1}{2z-1+z+2}=\frac{1}{2z-1}\frac{1}{1+\frac{z+2}{2z-1}}=\frac{1}{2z-1}\frac{1}{1-(-\frac{z+2}{2z-1})}=\frac{1}{2z-1}\sum_{n=0}^{\infty}(-\frac{z+2}{2z-1})^n=\frac{1}{2z-1}\sum_{n=0}^{\infty}(-1)^n\frac{(z+2)^n}{(2z-1)^n}=\sum_{n=0}^{\infty}(-1)^n\frac{(z+2)^n}{(2z-1)^{n+1}}$$?
 [/INST] No, your expansion is not a power series centered at $-2$. 
Let $w=z+2$ then for $|3w/5|<1$, or $|z+2|<5/3$,
\begin{align*}\frac{1}{3z+1}&=\frac{1}{3(w-2)+1}=\frac{1}{3w-5}=\frac{-1/5}{1-3w/5}=
-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3w}{5}\right)^n\\
&=-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3(z+2)}{5}\right)^n.
\end{align*}
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