For example, if we have:


Where the ratio test is satisfied. That is $\exists L$ s.t.

$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1$$

Does this mean that for the same $L$ that the root test's terms converge as well? That is, is it true that:

$$\lim_{n \to \infty} \left| a_n^{1/n} \right| = L$$

This is for a homework problem, but the homework is not to prove the general case; I was just wondering if this was true or not.


marked as duplicate by Alex M., user147263, Leucippus, Daniel W. Farlow, user228113 Mar 23 '16 at 0:39

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The answer to your question is yes: for any $L \in [0,\infty]$, if $\frac{a_{n+1}}{a_n} \rightarrow L$, then also $a_n^{\frac{1}{n}} \rightarrow L$.

Here is a stronger result:

Theorem: For any sequence $\{a_n\}_{n=1}^{\infty}$ of positive real numbers one has

$$\liminf_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$.

For a proof see e.g. $\S 5.3$ of these notes.

Note that the converse is not true: it is possible for the root test limit to exist but the ratio test limit not to. In fact we can get this just by mildly rearranging the terms of a convergent geometric sequence,e.g.

$\frac{1}{2}, \frac{1}{8}, \frac{1}{4}, \frac{1}{32}, \frac{1}{16} \ldots$

Here half of the successive ratios are $\frac{1}{4}$ and the other half are $2$, so the ratio test limit does not exist. But the root test limit is still $\frac{1}{2}$.

  • $\begingroup$ Yes. Thank you. I found that inequality shortly after posting this question. My experience with lim inf's and lim sup's is minimal, so correct me if I am wrong, but if the limit exists (for the ratio test) than lim inf = lim sup which would force the limit of the root test to exist and be the same as the ratio test's limit. $\endgroup$ – Stefan.Hannie Mar 20 '13 at 4:58
  • $\begingroup$ I think you meant $ \S 5.2\,$ , @Pete $\endgroup$ – DonAntonio Mar 20 '13 at 5:04
  • $\begingroup$ @Don: I think I do mean $\S 5.3$, since the result I quoted is Proposition 57 in that section. But anyway, close enough... $\endgroup$ – Pete L. Clark Mar 20 '13 at 5:09
  • $\begingroup$ @Stefan.Hannie: yes, that's exactly right. In fact, for the purpose of this question all you need to know about lim infs and lim sups is: (i) they exist for any sequence, but are possibly $\infty$ (or $-\infty$ if negative terms were allowed), and (ii) the limit exists (possibly $\pm \infty$) if and only if the lim inf and lim sup are equal. So the result you're asking about follows from this in a manner reminiscent of the Squeeze Theorem. $\endgroup$ – Pete L. Clark Mar 20 '13 at 5:12

The limit

$\lim_{n\to\infty}\sup\left|\frac{a_{n+1}}{a_n} \right| = L < 1$

is always greater than or equal to the limit

$\lim_{n\to\infty}\sup\left| a_n^{1/n} \right| = L$

So the root test is stronger than the ratio test. One can find cases when the root test shows convergence but the ratio test does not. In fact, the ratio test is a corollary of the root test. For example see: S. Krantz Real Analysis and Foundations, Chapman and Hall/CRC (Remark 4.1 on page 105, second edition of the book). It comes out directly from the proof of the ratio test.


Yes,you can show that by proving the following:

$$ if\quad a_n\ge 0\quad and\quad \lim_{n\to\infty}a_n=L,\quad then $$ $$ \lim_{n\to\infty}(\prod_{k=1}^{k=n} a_k)^{\frac{1}{n}}=L $$ Check by definition

  • $\begingroup$ What you wrote after that first "yes" is true yet it does not answer the question, which btw has a negative answer... $\endgroup$ – DonAntonio Mar 20 '13 at 4:50
  • $\begingroup$ @DonAntonio Do you believe the limit is also $L$? Since $a_n=a_0\cdot\prod_{k=1}^{k=n}\frac{a_k}{a_{k-1}}$, and $\lim_{n->\infty}\frac{a_n}{a_n-1}=L$. $\endgroup$ – Coiacy Mar 20 '13 at 4:53
  • $\begingroup$ I'm not following you: the claim you seem to be using is that if the limit of a positive sequence exists then both the limits of arithmetic andthe geometric means exist and are equal to the first limit. Fine. How though from this does it follow that $\,\sqrt[n]{|a_n|}\to L\,$ ,too? $\endgroup$ – DonAntonio Mar 20 '13 at 4:58
  • $\begingroup$ That is an exercise, check by definition using the $\epsilon$ and $N$ language. $\endgroup$ – Coiacy Mar 20 '13 at 5:01
  • $\begingroup$ Ah, ok...hehe. I thought that you confused both things. Perhaps you should make this clearer in your post. $\endgroup$ – DonAntonio Mar 20 '13 at 5:05

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