$\sum_\limits{n=1}^{\infty}(2-x)(2-x^{1/2})\cdots(2-x^{1/n}))$ convergence domain 
Exercise: Study the absolute and conditional convergence domain of the following series: $$\sum_\limits{n=1}^{\infty}(2-x)(2-x^{1/2})\cdots(2-x^{1/n}))\quad x>0$$

I have never seen a series like this. I tried to solve it on the following way:
$$\begin{align}\sum_\limits{n=1}^{\infty}(2-x)(2-x^{1/2})\cdots(2-x^{1/n}))&=\sum_\limits{n=1}^{\infty}\exp(\ln((2-x)(2-x^{1/2})\cdots(2-x^{1/n})))\\&=\sum_\limits{n=1}^{\infty}\exp(\ln(2-x)+\ln(2-x^{1/2})+\cdots+\ln(2-x^{1/n})))\\&=\sum_\limits{n=1}^{\infty}((2-x)+(2-x^{1/2})+\cdots+(2-x^{1/n})\\&=\sum_\limits{n=1}^{\infty}(2-x^{1/n})\end{align}$$
However $\lim_\limits{n\to\infty}(2-x^{1/n})=1\neq 0$ which implies the series do not converge.
Question:


*

*How do I deal with series like this? 

*Is my resolution right? If not, what is wrong?
 A: If $x$ is a positive power of $2$, it is in fact a finite sum. As noted by David, if $0<x\le 1$, then the series diverges. Let
$$
a_n=(2-x)(2-x^{1/2})\dots(2-x^{1/n}).
$$
Fix $x>1$ not a power of $2$. Let's apply the ratio test.
$$
\lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|=\lim_{n\to\infty}\bigl|2-x^{1/(n+1)}\bigr|=1.
$$
Since the test is inconclusive, let's try Rabbet's test. Observe that $2-x^{1/(n+1)}>0$ for $n$ large enough.
\begin{align}
\frac{a_n}{a_{n+1}}-1&=\frac{x^{1/(n+1)}-1}{2-x^{1/(n+1)}}\\
&=\frac{\log x}{n+1}+O(n^{-2}).
\end{align}
Then
$$
\lim_{n\to\infty}n\Bigl(\frac{a_n}{a_{n+1}}-1\Bigr)=\log x.
$$
The series converges if $x>e$ and diverges if $x<e$.
If $x=e$ then
$$
\frac{a_n}{a_{n+1}}-1=\frac{1}{n+1}+O(n^{-2})=\frac{1}{n}+O(n^{-2})
$$
and the series diverges by Gauss's test.
A: Edit: Since an infinitely superior answer has been given I thought maybe I should delete this. But that answer refers to this answer for one triviality; there's also a point below that's relevant to the other answer. Deleting the parts that are now moot:
Say $a_n=(2-x)(2-x^{1/2})\dots(2-x^{1/n})$. If $0<x\le1$ then $a_n\ge1$ for all $n$ so $\sum a_n$ diverges.
For $x>1$ see the other answer. One relevant technicality:
If $x>2$ it's not true that $a_n\ge0$, so strictly speaking Raabe's test cannot be applied. But in that case there exists $N$ so that $2-x^{1/N}\le0$ but $2-x^{1/(N+1)}>0$. Note that $\sum a_n$ converges if an only if $\sum_{n>N}a_n$ converges; for the latter sum start by factoring out the negative factors in each $a_n$ and apply Raabe on what remains.
