The given series can be rewritten as

$$\sum_{n\ge 1}\left(e^{\frac{\ln{n}}{n^s}}-1\right)$$

By the conclusion of # 4522, we have

$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$

Therefore, this series and the series $\displaystyle\sum_{n\ge 1}\frac{\ln{n}}{n^s}$ either both diverge or both converge. Therefore, it is sufficient to study $\displaystyle\sum_{n\ge 1}\frac{\ln{n}}{n^s}$.

When $s \le 1$, we have $\frac{\ln{n}}{n^s} > \frac{1}{n}$ for large $n$. Hence, the original series diverges.

When $s > 1$, we have $\ln {n} < n^{\frac{s-1}{2}}$ for large $n$. Therefore

$$\frac{\ln n}{n^s} < \frac{1}{n^{\frac{s+1}{2}}} < \frac{1}{n}$$

which means the original series converges.