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Both the ratio test and the root test define a number (via a limit).

If both limits exist (and shows that the series is convergent), what (if any) is the relation between the 2 numbers ? are they equal ? What is the relation (if any) between them and the original series (other than the fact that they say the series is convergent) ?

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marked as duplicate by Alex M., ᴡᴏʀᴅs, zz20s, Daniel W. Farlow, Jon Mark Perry Mar 23 at 0:45

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.

Page 3 here may help. – David Mitra Apr 7 '12 at 22:06
@DavidMitra - It is only 3 pages... – Belgi Apr 7 '12 at 22:07
Sorry, I meant page 3. – David Mitra Apr 7 '12 at 22:08
up vote 1 down vote accepted

For your first question:

If both limits exist, they must be equal to each other. In fact, for a sequence of positive terms $(a_n)$, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}\root n \of {a_n}$ and moreover, in this case, the two limits are equal to each other. This follows from a more general fact contained in these notes of Pete L. Clark.

I'm not sure if the following answers your second question, but:

In general, there is no relationship between the value of the limit $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ and the value of the sum $\sum\limits_{n=1}^\infty a_n$.
Indeed, here is a silly example showing this:

Suppose $(a_n)$ is a sequence of positive terms and that $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$. Then $\sum\limits_{n=1}^\infty a_n$ converges, say to $S\ne 0$. Now let $a>0$ and consider the sequence $(b_n)$ defined by $b_n=a\cdot a_n$. Here we have $\lim\limits_{n\rightarrow\infty} {b_{n+1}\over b_n} =\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}= r$. But, $\sum\limits_{n=1}^\infty b_n=aS$.

So if, $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$, the corresponding series could possibly converge to any given positive number. The same remark holds for the limit in the Root test.

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For a non-negative real series $(a_n)_{n \in \mathbb{N}}$, the tests give two (possibly undefined) numbers: let’s call them $L_\textit{root} := \lim_n (a_n)^{\frac{1}{n}}$, and $L_\textit{ratio} := \lim_n \frac{a_{n+1}}{a_n}$.

From Lemma 3 of these notes by Pete L. Clark, it follows that if $L_{\textit{ratio}}$ is defined, then $L_\textit{root}$ is also defined, and they are equal.

This is reasonably intuitive, with a bit of thought: suppose that for $n>N$, the ratio of consecutive terms $\frac{a_{n+1}}{a_n}$ is always close to $L$. Then (still for $n>N$), consider $a_n$ as produced by multiplying $a_N$ by all the later consecutive ratios; so it’s close to $L^{n-N} a_N$, and its $n$th root is close to $(L^{n-N} a_N)^{\frac{1}{n}} = L (\frac{a_N}{L^N})^\frac{1}{n}$. The second factor here, being the $n$th root of a constant, goes to $1$ as $n$ grows; so for sufficiently large $n$, $(a_n)^\frac{1}{n}$ will be close to $L$. (Exercise: make this argument precise — replace each “…close to…” by appropriate specific bounds.)

On the other hand, the converse doesn’t generally hold. $L_\textit{root}$ may be defined even if $L_\textit{ratio}$ is not. For example, set $a_n = 2^n$ when $n$ is even, $a_n = 2^{n-1}$ when $n$ is odd. Then the ratio of consecutive terms alternates between 1 and 4, so $L_\textit{ratio}$ is undefined; but the sequence is close enough to $2^n$ that the root converges, with $L_\textit{root} = 2$.

(Thanks to @David Mitra’s comment for the reference to the linked notes.)

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Both tests will yield the same property about the series:

  1. if $L<1$ the series is absolutely convergent
  2. if $L>1$ the series is divergent.
  3. if $L=1$ the series may be divergent, conditionally convergent, absolutely convergent

Edit: Note that if L=1 in the ratio test, the root test will also yield L=1. The converse is not true. (I had mistakingly asserted that it was earlier.)

As for relation... I'm not sure what you mean. The root test can be considered more comprehensive as it yields information whenever the ratio test is inconclusive. Applying the ratio test, however, can simpler in certain cases or perhaps necessary like when dealing with factorials.

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Are you saying it's impossible that $L=1$ for one test and $L=\text{some other positive number}$ for the other test? How would you know that? – Michael Hardy Apr 7 '12 at 22:26
Yes. It's apparent when comparing the proofs. I recommend reading my proof alongside tests' proofs: Divergence: If $L>1,$ Ǝ $N: n\geq N$ yeilding $|a_{n+1}|/|a_n|>1⇒|a_{n+1}|>a_n$ and $|a_n|^{1/n}>1$⇒$|a_n|>1$. For both cases, $\lim_{n\to\infty}a_n\neq 0$ the test fails iff $L>1$ and $a_n \neq 0$ . Convergence: Ǝ $q,r: L<q<1$, $|a_{n+1}|<q|a_n|$⇒$|a_{n}|q⇒|a_{N+k}|<q|a_{N+k-1}<q^{k}|a_N|$ and $|a_n|<r^n$. The tests continue to show that the r,q series geometric nature shows both series converge by the comparison test. Therefore, the tests succeed iff $L<r,q<1$. – faraz Apr 8 '12 at 21:41
I find your remarks unclear. For some divergent series, one has $L=1$ for at least one of the tests. You're saying if that happens, then $L$ must be $1$ for both tests. Why? Likewise, for some convergent series, one has $L=1$ for at least one of the tests. And again, you're saying in that case it must happen for both tests. Can you explain why? – Michael Hardy Apr 9 '12 at 1:43
Both tests reduce to the same "subtests" after inconsequential manipulation of the series. When $L>1$, the reduction leads to identical divergence tests: $\lim{a_n} \neq 0$ or DNE. Divergence proof for root test: Given $L = \lim_{n \to \infty}|a_n|^{1/n}$, when $L>1$ Ǝ $N: n\geq N ∴ |a_n|^{1/n}>1 ⇒ |a_n|>1 ∴ \lim_{n \to \infty}a_n≠0$ (divergence test.) Divergence proof for ratio test: Given $L = \lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|$, when $L>1$ Ǝ $N: n\geq N ∴ |\frac{a_{n+1}}{a_n}| > 1 ⇒ |a_{n+1}| > |a_n| ∴ \lim_{n \to \infty}a_n≠0.$ (Identical divergence test used to prove the root test.) – faraz Apr 9 '12 at 3:01
Root test: $L=\lim_{n\to\infty}|a_n|^{1/n}$ Ǝ $r: L<r<1$ and since Ǝ $n: n\geq N, |a_n|<r^n$. Since $1..r^n$ is a geometric series by the comparison test, $a_N..a_n$ converges and $a_1..a_N$ is finite. Ratio test: Given $L = \lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|$, when $L<1$ Ǝ $r: L<r<1$ and since Ǝ $n: n\geq N, |a_{n+1}|<r|a_n|$. Consider $|a_{N+k}|<r|a_{N+k-1}|<r^k|a_N|$. Let $N+k$ by represented by $n$, $|a_n|<r^{n-N}|a_N|$ or $|a_n|^{1/n}<r^{(n-N)/n}|a_N|^{1/n}$ which is a reincarnation of the root test. – faraz Apr 9 '12 at 4:06

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