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?