Questionable Power Series for $1/x$ about $x=0$

Question

WolframAlpha states that

The power series for $1/x$ about $x=0$ is:

$$1/x = \sum_{n=0}^{\infty} (-1)^n(x-1)^n$$

This is supposedly incorrect, isnt it?

This is showing the power series about $x=1$ in the form $(x - c)$

I dont understand how WolFramalpha says that is correct:

http://m.wolframalpha.com/input/?i=power+series+of+1%2Fx&x=0&y=0

Thanks!

Answer 1

You may recall that $$ \sum_{k=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|<1. \tag1 $$ Just substitute $r \rightarrow 1-x $ in $(1)$, you get, for the right hand side $$ \frac{1}{1-r}=\frac{1}{1-(1-x)}=\frac 1 x $$ as long as $|1-x|<1$ ($0$ being excluded).

Answer 2

It is rather strange, because Wolfram Alpha is perfectly happy to return a Laurent series for e.g.

series of 1/(x+x^2) at x = 0

Somehow, $1/x$ is treated differently.