I'm asked to look at $$\sum_{n=1}^{\infty}{\frac{x^{-n}}{\sqrt{n}}}$$ and decide whether it's a fine power series, and if so: then around which point it's developed. Since I know a power series is of the form: $$\sum_{n=0}^{\infty}{a_n(x-x_0)^n}$$ I can say right away that: $$a_n = \frac{1}{\sqrt{n}}$$ All of the examples I've seen until today never included $x$ (or $x^n$ for that matter) at the denominator. Suppose such form exists, I can assume the series is developed around $x_0=0$. And here's my dillemma:
Suppose such series does exist, hence the domain of $x$ so that this power series converges is $|x|>1$, since our $x$ is in the denominator. Can that series still be considered valid if it's developed around $x=0$ but $x$ has to be greater than 1?
From all the $x$'s in $\mathbb{R}$, $x=0$ is the only value that is a singular point. Since this is a singular point and the function to which this series converges to is not analytic at $x=0$ can this series can still be called a valid power series? (would it be a valid one if it were: $\sum_{n=1}^{\infty}{\frac{(x-3)^{-n}}{\sqrt{n}}}$ for example?).
I would appriciate it if someone could clarify it for me.
This is a perfectly valid Laurent series (which is a generalization of power series) with inner radius of convergence $1$ and outer radius of convergence $\infty$; in other words, it converges for $|x|>1$ (and also, as it happens, at $x=-1$).