Question 14 - Examen 2019
La série numérique \(\displaystyle \sum_{n=1}^\infty \frac{1}{\sqrt[5]{n^{\frac{2}{\alpha}}\bigl(n^{2\alpha} +1\bigr)}} \) converge si
  • \(1\lt \alpha\lt 2\)
  • \(\alpha=\frac12\)
  • \(0\lt \alpha\lt \frac{1}{2}\)
  • \(\frac12\lt \alpha\lt 1\)