2

My question is motivated from this question.

Is converse true? Does the convergence of the root test imply convergence under the ratio test for sequences? Most textbooks covering series claim that the root test is stronger than the ratio test, so I suspect the converse should also be true. Is it?

Community
  • 1
Junaid Aftab
  • 1,582

1 Answers1

1

No, it's not true. The fact that $\lim_{n\to\infty}\sqrt[n]{|a_n|}$ exists does not imply that $\lim_{n\to\infty}\big|\frac{a_{n+1}}{a_n}\big|$ exists.

For example, take $a_1=1$ and $a_{n+1}=a_n/2$ if $n$ is not a perfect square, but $a_{n+1}=2a_n$ if $n$ is a perfect square. The root test gives a limit of $1/2$, so the series is convergent, but the ratio test does not give a limit (since the ratio is $1/2$ infinitely often and $2$ infinitely often), so is inconclusive.

Especially Lime
  • 40,828
  • Oh, I see your point. But when most textbooks mention that the root test is more strong, I naturally think...this means convergence under the root test also implies convergence in the ratio test. Where am I going wrong? – Junaid Aftab Mar 18 '17 at 18:01
  • No, it's the other way around: convergence for the ratio test implies convergence for the root test. So whenever the ratio test works (i.e. tells you whether the series converges), the root test also works, but sometimes the root test works even when the ratio test doesn't. – Especially Lime Mar 18 '17 at 18:10
  • A stronger statement implies a weaker one, but it's not the same with tests. A stronger test is one that works more often, but a stronger statement is one that's true less often. – Especially Lime Mar 18 '17 at 18:14