I am told to find the error in the following argument:
let $f(z)= \cdots + \frac{1}{z^2} + \frac{1}{z} + 1 + z + z^2 + \cdots$
Note that $z + z^2 + \cdots = \frac{z}{1-z}$
and $1 + \frac{1}{z} + \frac{1}{z^2} + \cdots = \frac{-z}{1-z}$
hence f(z) = 0
My argument is to point out that $z + z^2 + \cdots = \frac{z}{1-z}$ only holds when |z| < 1 and $1 + \frac{1}{z} + \frac{1}{z^2} + \cdots = \frac{-z}{1-z}$ can only hold in the region where 1 < |z|. Is this a sufficient counterargument or is there something else I am missing because this argument seems too simple.
thank you
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.