Divergent or convergent but how ??

Question

I was to depict the convergence & divergence nature of the summation $\sum A_n$ where $A_n = (n^{1/n}-1)^k$ I was able to prove that when $k>1$ then $\sum A_n$ is converging and while $k<0$ it was diverging but when $1>k>0$ I saw the sequence $A_n$ to be increasing but I am unable to prove if the summation $\sum A_n$ is converging or diverging, so how do I prove it ?

Answer 1

Since $e^x\geq x+1$,

$$ n^{\frac{1}{n}}-1 = \exp\left(\frac{\log n}{n}\right)-1 \geq \frac{\log n}{n} $$ hence if $0<\alpha\leq 1$, $$ \sum_{n=1}^{N}\left(n^{1/n}-1\right)^\alpha \geq \sum_{n=1}^{N}\frac{\log(n)^\alpha}{n^\alpha} $$ and the last series is diverging as $N\to +\infty$ by comparison with the generalized harmonic series.