I am looking to understand why my construction of the Laurent series for the following function does not seem correct.
Consider the function $f(z) = z^{-1} + (z + 1)^{-1} + (z + 2)^{-1}$ on the annulus $\{z\mid 0 < |z| < 1\}$. I understand that one quick way to determine the Laurent series for this function, and for any function that is a linear combination of terms involving $z$, is to first construct the Laurent series for $(z + 1)^{-1}$ and $(z + 2)^{-1}$ and then consider their sum with $z^{-1}$.
I decided to try to construct the same series with the (somewhat painful) general form of the Laurent series, namely $f(z) = \sum_{n=-\infty}^\infty a_n(z - 0)^n$ with $a_n = \frac{1}{2\pi i}\int_{\partial B_r(0)}\frac{f(y)dy}{(y - 0)^{n+1}} = \frac{1}{2\pi i}\int_{\partial B_r(0)}\left(y^{-1} + (y + 1)^{-1} + (y + 2)^{-1}\right)y^{-(n+1)}dy$ where $0 < r < 1$.
One can use partial fraction decomposition to write
$$(y + 1)^{-1}y^{-(n+1)} = 1\cdot (y + 1) + (-1)^{n+1}y^{-(n+1)}$$
$$(y + 2)^{-1}y^{-(n+1)} = \frac{1}{2}\cdot (y + 1) + \left(\frac{-1}{2}\right)^{n+1}y^{-(n+1)}$$
Thus,
$$a_n = \frac{1}{2\pi i}\int_{\partial B_r(0)}\left(y^{-(n+2)} + y^{-(n+1)} + (-1)^{n+1}(y + 1)^{-1} + \frac{1}{2}y^{-(n+1)} + \left(\frac{-1}{2}\right)^{n+1}(y+2)^{-1}\right)dy$$
Since $(y+1)^{-1}$ and $(y + 2)^{-1}$ are analytic on $B_r(0)$ for any $0 < r < 1$, their respective integrals vanish. Therefore the integral reduces to
$$a_n = \frac{1}{2\pi i}\int_{\partial B_r(0)}\left(y^{-(n+2)} + \frac{3}{2}y^{-(n+1)}\right)dy$$
Since we are integrating over the boundary of the open disk $B_r(0)$, $y = re^{i\theta}, \theta \in [0, 2\pi)$. Thus,
$$a_n = \frac{1}{2\pi}\int_{[0,2\pi)}\left(\left(r\exp(i\theta)\right)^{-(n+1)} + \frac{3}{2}\left(r\exp(i\theta)\right)^{-(n)}\right)d\theta$$
(Possible source of error) Since $\exp$ is an entire function and we are integrating over a closed path we have that $a_n = 0$ for $n \not\in \{-1, 0\}$, as only for those $n$ does the $\exp$ term vanish. But this then implies that $f(z) = \frac{3}{2} + \frac{1}{z}$ which differs quite a bit from $f(z) = \frac{1}{z} + \sum_{n=0}^\infty(-1)^n\left(1 - 2^{-(n+1)}\right)z^n$, a series which I obtain with the first approach.
So where does my reasoning fail?
More simply:
Infinite Geometric series is given as: $(1+u)^{-1}=1-u+u^2-u^3+.....$ if $|u|<1.$
So when $0<|z|<1$, we can write $$(1+z)^{-1}=1+z+z^2+z^3+....(1)$$ and $$(2+z)^{-1}=2^{-1}(1+z/2)^{-1}=2^{-1}[1-(z/2)+(z/2)^2-(z/3)^3+...]....(2).$$