Proof that $\sum_{n = 1}^\infty\bigl(\sum_{k = 1}^\infty 1 / n^k\bigr)$ doesn't converge

Question

How to prove that $\sum_{n = 1}^\infty\Bigl(\sum_{k = 1}^\infty \frac 1 {n^k}\Bigr)$ doesn't converge as wolfram said ?

Answer 1

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^k} \ge \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{n^k}\ge \sum_{n=1}^\infty \frac1n$$

As the harmonic series, $\sum_{n=1}^\infty \frac1n$, is divergent, it follows that the double summation is divergent.

Answer 2

$$\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{n^k} = \sum_{n=1}^\infty \frac{1}{n} + \sum_{k=2}^\infty \sum_{n=1}^\infty \frac{1}{n^k}$$

The sum $\sum_{n=1}^\infty \frac{1}{n}$ diverges, so the whole thing diverges.