Does the series $\sum\limits_n(2^{1/n} - 1)$ converge? Does the series $\sum\limits_{n=1}^\infty (2^{1/n} - 1)=(2^1 - 1) + (2^{\frac{1}{2}} - 1)+ ... +(2^{\frac{1}{n}}-1)+...$ converge?

I feel like it diverges the same way as $\sum \frac{1}{n}$ diverges: very slow, on a logarithmic scale.
Your intuition is correct!

Using the fact that $e^x-1 \ge x$ for all real $x$, we have $2^{\tfrac{1}{n}}-1 = e^{\tfrac{1}{n}\ln 2}-1 \ge \dfrac{1}{n}\ln 2$ for all $n$. 

Then, since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n}\ln 2$ diverges, $\displaystyle\sum_{n = 1}^{\infty}2^{\tfrac{1}{n}}-1$ diverges by direct comparison.
You can also go about this problem by using the Comparison Test:

Let $\sum$ $a_n$, $\sum$ $b_n$ be 2 positive sequences where $a_n$ $\le$ $b_n$. $$\\$$
$\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ $\ge$ $\sum\limits_{n=1}^\infty (-1)$

Since every infinite sum of a non-zero constant diverges, $\sum\limits_{n=1}^\infty (-1)$ diverges. Thus $\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ diverges.