When $n$ is large, $$\frac{1}{\sqrt[n]{\log n}}\simeq 1-\frac{\log (\log (n))}{n}$$ So, as you already noticed $$\lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{\log n}}=1$$ and then the sum $$\sum_{n=0}^\infty \frac{1}{\sqrt[n]{\log n}}$$ diverges.
As said in other answers, the summation should probably start at $n=2$ and not $n=0$ ($n=0$ does not make a problem but $n=1$ makes a serious one).
Using some approximations, it could be easy to show that, for $p \gt 1000$, $$\sum_{n=2}^{n=p} \frac{1}{\sqrt[n]{\log n}}\simeq p$$