Does the series sum n=0 to oo ( 2^(1/n) - 1 ) converge?

I'll assume that you mean Σ(n = 1 to ∞) (2^(1/n) - 1).

----

Use the Limit Comparison Test with the harmonic series

Σ(n = 1 to ∞) 1/n.

lim(n→∞) (2^(1/n) - 1)/(1/n)

= lim(t→0+) (2^t - 1)/t, letting t = 1/n

= lim(t→0+) (2^t ln 2 - 0)/1, by L'Hopital's Rule (0/0 version)

= ln 2.

Since ln 2 is a nonzero constant and Σ(n = 1 to ∞) 1/n is a known divergent series, we conclude that Σ(n = 1 to ∞) (2^(1/n) - 1) also diverges by the Limit Comparison Test.

I hope this helps!

It diverges. 2^(1/n) is always greater than 1, so 2^(1/n) - 1 is always greater than 0. That's not a rigorous reason, I know, but the rate at which 2^(1/n) - 1 tends to 0 is so slow that the infinite sum diverges

Yes. When n is very large 2^(1/n) tends to 1. The limit goes to o. But n=0 is a singular point. If it is included then the series tend to infinity.