When the series $\sum_{n=2}^{+\infty }(\sqrt[n]{n}-1)^a$ is convergent depending on $a\in\Bbb R$?
For $a=0 $ the series is divergent.
Else:
$\sqrt[n]{n}-1=\frac{e^{1/n\ln n}-1}{1/n \ln n}\cdot \frac{1}{n}\ln n\sim\frac{\ln n}{n}$
So, $\sqrt[n]{n}-1$ is asymptotically similar to $\frac{\ln n}{n}$, and $(\sqrt[n]{n}-1)^a$ will converge when $(\frac{\ln n}{n})^a$ converges. But I don't know how to finish it. Could you help me?
On
In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$\sqrt[n]{n}-1\le \frac{e}{e-1} \frac{\log(n)}{n}$". To address the OP's request, we now proceed.
Using $e^x<\frac1{1-x}$ for $x<1$ and $\min\left(1-\frac1n\log(n)\right)=1-e^{-1} $ we have
$$\begin{align} \sqrt[n]{n}-1 &=e^{\frac1n\log(n)}-1\\\\ &\le \frac{\frac1n\log(n)}{1-\frac1n\log(n)}\\\\ &\le \frac{e}{e-1} \frac{\log(n)}{n}\end{align}$$
For $a>1$, compare to $\frac1{n^{1+(a-1)/2}}$: $$\lim_{n\to\infty}\left(\frac{\ln n}{n}\right)^a\Big/\frac1{n^{1+(a-1)/2}}\\ =\lim_{n\to\infty}\frac{\ln^an}{n^{a/2-1/2}}$$ Since $a>1$, $\frac a2-\frac12>0$, hence the limit equals $0$. The series converges.
For $a\le 1$, compare to $\frac1n$:$$\lim_{n\to\infty}\left(\frac{\ln n}{n}\right)^a\Big/\frac1{n}\\ =\lim_{n\to\infty}\frac{\ln^an}{n^{a-1}}=\infty$$ Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.