2

A professor once told that if limit of th ratio/root test is a finite L then the limit of the other test yields the same number L. Can someone confirm this? in which cases can I say that both tests yield the same limit? Thanks.

Daniel Turizo
  • 325
  • 3
    The root test is strictly stronger. We have (assuming everything nonzero) $$\liminf_{n\to\infty} \frac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert} \leqslant \liminf_{n\to\infty} \lvert a_n\rvert^{1/n} \leqslant \limsup_{n\to\infty} \lvert a_n\rvert^{1/n} \leqslant \limsup_{n\to\infty} \frac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert},$$ so if the limit of the ratios exists, the limit of the roots also exists, and they are the same. But the limit of the roots can exist without the limit of the ratios existing. – Daniel Fischer Aug 02 '14 at 17:50
  • See also here. – Daniel Fischer Aug 02 '14 at 17:53
  • http://math.stackexchange.com/questions/814500/is-lim-k-to-infty-x-k-frac-1k-r-true/814514#814514 – Swapnil Tripathi Aug 02 '14 at 18:00
  • An example of a series for which the Ratio test fails but the Root test does not is $\sum_{n=1}^{\infty}1/(2^{n+(-1)^{n}})$ – user84413 Aug 02 '14 at 22:29
  • I thank you @DanielFischer. I only have one doubt left: if the limit of the ratios is infinite, is the limit of the roots infinite too? – Daniel Turizo Aug 02 '14 at 23:22
  • Yes, the inequalities above hold also for infinite $\liminf$, so $\frac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert}\to \infty \implies \lvert a_n\rvert^{1/n} \to \infty$. – Daniel Fischer Aug 02 '14 at 23:27
  • Thank you again! It's all clear to me now. – Daniel Turizo Aug 04 '14 at 22:54

0 Answers0