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
$$\lim S_k= \infty$$
Best Answer
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 $\sum 2^{-n}$.