Is $\sum (n^{\frac{1}{n}}-1)^n$ divergent or convergent?

Since $n^{\frac{1}{n}}\geq 1$ for $n\geq 1$ and $\lim_{n\to\infty}n^{\frac{1}{n}}=1$, there exists some $N$ such that $$ 1\leq n^{\frac{1}{n}}\leq \frac{3}{2} $$ for all $n\geq N$ (in fact we can take $N=1$, but this isn't so important.)

Therefore $$ 0\leq (n^{\frac{1}{n}}-1)^n\leq \Big(\frac{3}{2}-1\Big)^n=2^{-n}$$ for all $n\geq N$, so your series converges by comparison with $\sum 2^{-n}$.

It's hard to follow your line of reasoning. We cannot say that "because $(k^{1/k} - 1)^k \to 0$, the series is divergent". In fact, in order for the series to be convergent (opposite of divergent), we must have $(k^{1/k} - 1)^k \to 0$.

For this series, the root test is particularly useful: $$ \lim_{n \to \infty}[(n^{1/n} - 1)^n]^{1/n} = \lim_{n \to \infty}(n^{1/n} - 1) = 0 < 1 $$ We conclude that the series must converge.

Let $0<\varepsilon <1$. Since $\lim_{n\to \infty}n^{1/n}=1$, as you stated, we have $\lim_{n\to \infty} n^{1/n}-1=0$, so there exists $N\in \mathbb N$such that whenever $n>N$ we have $0<n^{1/n}-1<\varepsilon$. Therefore, for such $n$, we have $0<(n^{1/n}-1)^n<\varepsilon^n$. Consequently, by the comparison test, since $$\sum_{k=1}^\infty \varepsilon^k$$ converges, so does the original sum.

It's not the case that the sum of an infinite number of positive terms is infinite. For example, $$\frac 1 2+\frac 1 4+\frac 1 8 +... = 1$$