<s>[INST] (complex variables) Expand $\frac{2z+3}{1+z}$ in a power series of $z-1$ and comment on its convergence Question:
Expand $\frac{2z+3}{1+z}$ in a power series of $z-1$. What can we say about its convergence?
Attempt:
First, notice $ \frac{2z+3}{1+z} = \frac{2z+3}{1} \frac{1}{1+z}$.
Let $w = 1 -z$. Using substitution, I can get
$$ \frac{5}{2} - \sum_{n=1}^\infty (z-1)^n (-1)^n 2^{-n}.$$
So, it converges when $ \left| z - 1 \right| < 1$.
However, if I plug this into wolfram alpha, I get:
$$ \frac{5}{2} + \sum_{n=1}^\infty (z-1)^n (-1)^n 2^{-1-n}.$$
Notice that in the second case, the summation is added to the fraction and there is an additional $\frac{1}{2}$ in the summation. Can anyone tell me where I'm going wrong?
 [/INST] Let $w=z-1$ so
$$\frac{2z+3}{1+z}=\frac{2w+5}{w+2}=2+\frac{1}{w+2}=2+\frac12.\frac{1}{1+\frac{w}{2}}=2+\frac12\sum_{n=0}^\infty(-1)^n\Big(\frac{w}{2}\Big)^n$$
and this series with convergence redius 2 expand as $$\frac52-\frac{w}{4}+\frac{w^2}{8}-\frac{w^3}{16}+\cdots$$
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