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

I tried to follow the idea of $k^{th}$ partial sum from another question that I asked: Verify if $\sum(\sqrt{n+1}-\sqrt{n})$ is convergent or divergent
$$S_k=\sum_{n=1}^k (n^\frac{1}{n}-1)^n=(1-1)+(2^\frac{1}{2}-1)^2+\dots+(k^\frac{1}{k}-1)^k$$
I know that $k^\frac{1}{k}\rightarrow 1$ when $k\rightarrow \infty$ and then $(k^\frac{1}{k}-1)^k\rightarrow 0$. So I have an infinite sum of small terms, what makes me think that this series is divergent. Since a sum of infinite terms bigger than $0$ is infinite
$$\lim S_k= \infty$$
 A: 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.
A: 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,
$$\frac 1 2+\frac 1 4+\frac 1 8 +... = 1$$
A: 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 $\sum 2^{-n}$.
