Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am a little bit confused regarding the meaning of the phrase :" Root test is stronger than ratio test", and was hoping you will be able to help me figure it out. As far as I can see here: The limit from the ratio test is greater or equal the limit from the root test . So, my first question is- is there any example of a series $\Sigma a_n$ such that the limit from the ratio test is exactly 1 (i.e.- inconclusive), but the limit from the root test is less than 1? (i.e.- convergence can be proved by using the root test but not by using the ratio test ) If not, then is it correct that this phrase is the meaning of "stronger" is when the limit from the ratio test does not exist? (as in the classic example of a rearranged geometric series)

Hope you will be able to help.

THanks !

related posts: Show root test is stronger than ratio test

Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$

Do the sequences from the ratio and root tests converge to the same limit?

share|cite|improve this question

marked as duplicate by Alex M., ᴡᴏʀᴅs, Daniel W. Farlow, G. Sassatelli, Jon Mark Perry Mar 23 at 1:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 2 down vote accepted

Consider the example of series

$$\sum 3^{-n-(-1)^n}$$

root test establishs the convergance but ratio test fails

onother example series with nth term $a_n=2^{-n}$ if n is odd $a_n=2^{-n+2}$ if n is even

for second series when n is odd or even and tends to $\infty$ ${a_n}^{\frac{1}{n}}=\frac{1}{2}$ Hence by cauchys root test the series converges but the ratio test gives $\frac{a_n}{a_n+1}=\frac{1}{2}$ if n is odd and tends to $\infty$

$\frac{a_n}{a_n+1}=8$ when n is even and approachs $\infty$ Hence ratio test fails.. Sorry I dnt know mathjax that is why i was a bit late...

share|cite|improve this answer
Will you please give some details ? (i.e.- why the root test establishes the convergence and the ratio test does not ? Thanks) – yehushua Mar 22 at 11:57

Root test is stronger in the sense $\exists\lim$ of quotient $\implies\exists\lim$ of root. When both limits exist, they are equal.

share|cite|improve this answer
Doesn't this mean the root test is at least as strong as the ratio test? The question asked for an example demonstrating that the root test is strictly stronger. – mb7744 Mar 22 at 18:20
@mb7744, the OP asked the sense of "stronger" and more concretely if it is possible that "... the limit from the ratio test is exactly 1 (i.e.- inconclusive), but the limit from the root test is less than 1". This is impossible. – Martín-Blas Pérez Pinilla Mar 22 at 19:06

Your last sentence is exact: note that if $${\lim \inf}_{n\to\infty} \frac{a_{n+1}}{a_n} \neq {\lim \sup}_{n\to\infty} \frac{a_{n+a}}{a_n} $$ then the limit does not exist. If otherwise $${\lim \inf}_{n\to\infty} \frac{a_{n+1}}{a_n} = {\lim \sup}_{n\to\infty} \frac{a_{n+1}}{a_n} $$ then the limit of the ratio test exists, so does the one of the root test, and both coincide.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.