[tex]\displaystyle Observam~ca~\frac{k^3+1}{k^4+1} \ge \frac{k^3}{k^4}.~Intr-adevar,~aceasta~este~echivalenta \\ \\ cu~k^7+k^4 \ge k^7+k^3 \Leftrightarrow k^4 \ge k^3,~adevarat. \\ \\ Atunci~\sum\limits_{k=1}^n \frac{k^3+1}{k^4+1} \ \textgreater \ \sum\limits_{k=1}^n \frac{k^3}{k^4}= \sum\limits_{k=1}^n \frac{1}{k} \to \infty.[/tex]
