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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: https://www.maa.org/sites/default/files/0025570x33450.di021200.02p0190s.pdf 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?

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yehushua
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3 Answers3

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

Martín-Blas Pérez Pinilla
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    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 '16 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 '16 at 19:06
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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...

Sine of the Time
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Rayees Ahmad
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    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 '16 at 11:57
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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.

Miguel
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