Return to Answer

The OP asks the question [now deleted] whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges.

The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.

The OP asks the question whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges.

The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.

The OP asks the question whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges.

The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.

The OP asks the question [now deleted] whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges.

The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.

Your first question has an obvious answer: The series $S$ diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.

The OP asks the question whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges.

The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.