To find out for which $\alpha>0$ the series $\sum_{k=1}^\infty\,\frac1{k^\alpha}$ converges, one can use the following inequality which is valid for all $n\in\llbracket2,\infty\llbracket$,

$\int_2^n\frac{{\rm d}x}{x^\alpha}\;<\;\sum_{k=2}^n\,\frac1{k^\alpha}\;<\;\int_1^n\frac{{\rm d}x}{x^\alpha}\:.$

This inequality is easily proved (for example with the help of a drawing) and implies that

\begin{align*} \frac1{1-\alpha}\left(\frac1{n^{\alpha-1}}-\frac1{2^{\alpha-1}}\right)\;&<\;\sum_{k=2}^n\,\frac1{k^\alpha}\;<\;\frac1{1-\alpha}\left(\frac1{n^{\alpha-1}}-1\right)\qquad\text{ for }\alpha\neq1\:,\\ \ln (n) - \ln (2)\;&<\;\sum_{k=2}^n\,\frac1{k}\;<\;\ln (n) - \ln (1)\:. \end{align*}

By letting n go to infinity we conclude that the series converges if $\alpha>1$ and diverges to infinity if $\alpha\leq1\,.$ Having this in mind, the following question is natural.

Exercise:

Let $\sum_{k=1}^\infty\,u_k$ be a convergent series. Is it true that the series $\sum_{k=1}^\infty\,u_k^3$ converges too?