Checking for convergence/divergence of the nth root. I've been working on this problem for a while, but I can't really get it. 
I get it, but I don't actually get it. 

The question is to find whether or not this series converges:
$\displaystyle\sum_{n=1}^\infty(2 ^{1/n} - 1) $

I am almost certain that it diverges, but the way I did it proved convergence, and looking into it online didn't give results I could understand, but generally agreed that it was divergent.

Let $f(x) = 2^{1/x} - 1$

The way I did it was using a comparison test of a p series $1/x^{1.1}$. Since I know $1/x^{1.1}$ is greater than $f(x)$ (this is apparently wrong, as a source online claims that $2^{1/x} - 1 \ge 1/n$) and because $p&gt;1, 1/x^{1.1}$ converges, which would therefore, by the limit comparison test, mean that $f(x)$ must also converge since it's less than $1/x^{1.1}$.

I am at a disagreement with $1/x$ being less than or equal to $f(x)$ because when I graphed it, it was greater. $1/x^{1.1}$ was also greater when graphed.

If anyone can shed light on how $1/x$ is less than or equal to $f(x)$, that'd be great, since then by that comparison if $1/x\ge f(x)$, $f(x)$ would also diverge.
Notice that

$$2^{1/n}-1=\exp\left(\frac1n\ln2\right)-1\sim_\infty\frac{\ln2}{n}$$
 so the series is divergent by comparison with the harmonic series $\sum\frac1n$.
Show the terms don't even approach zero (they approach 1) as n goes to infinity.  Then I think it's just usually called the divergence test that is applied.