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Proof

(related to Proposition: Limit Test for Roots or Ratios)

• By hypothesis, the sequence of n-th roots $(\sqrt[n]{|a_n|})_{n\in\mathbb N},$ or the sequence of ratios $\left(\frac{|a_{n+1}|}{|a_n|}\right)_{n\in\mathbb N}$ is convergent to a limit $\alpha > 0.$
• Case a) $\lim_{n\to\infty} \sqrt[n]{|a_n|}=\alpha < 1.$
  • By the inequality of arithmetic mean, we have $\alpha < (\alpha+1)/2 < 1.$
  • Therefore, there is an index $N$ such that for all $n > N$ we have $\sqrt[n]{|a_n|} < q:=(\alpha+1)/2.$
  • By the root test, the infinite series $\sum_{n=0}^\infty a_n$ is convergent.
• Case b) $\lim_{n\to\infty} \sqrt[n]{|a_n|}=\alpha > 1.$
  • As for the case a) $\alpha < 1$, the root test reveals that $\sum_{n=0}^\infty a_n$ is divergent, if $\alpha > 1.$
• Cases c) $\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}=\alpha < 1$ and d) $\alpha > 1.$
  • The reasoning is identical as in the cases a) and b), but we have to apply the ratio test.
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References

Bibliography

1. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition