# Ratio test and root test

In Mathematical Analysis by Tom M. Apostol it's stated that root test is "stronger" than ratio test because there are series whose convergence can be decided by the former, but not by the latter.

Question: I'm looking for an example of this. I prefer an example of a series of positive terms, where both limits of $$a_{n+1}/a_n$$ and $$\sqrt[n]{a_n}$$ exist. In other words, I need a sequence $$(a_n)_n$$ of positive terms such that $$a_{n+1}/a_n\to 1$$ and $$\sqrt[n]{a_n}\to r\neq 1$$.

Try: I have tried $$\sum\dfrac{n!}{n^n}$$, but the ratio test yields $$1/e<1$$. Any suggestion?

Note: The example may be divergent or convergent.

• – lhf
Commented Jun 7, 2021 at 18:53
• @lhf Thanks, but as I said, I prefer examples where both limits exist. The reader of the example has not the notion of $\limsup$. Commented Jun 7, 2021 at 18:57
• When both limits exist, they coincide.
– lhf
Commented Jun 7, 2021 at 19:18
• Note that (for positive terms) if the root test gives a limit which exists and is less than $1$ then the series does converge. So your "example may be divergent or convergent" should really only have "convergent" as an option. Commented Jun 7, 2021 at 19:18
• @coffeemath That's true. Edited. Commented Jun 7, 2021 at 19:22

Let $$(b_n)$$ be a sequence such that $$b_n>0$$ for every $$n$$ and $$\ell>0$$. If $$\lim_{n\to\infty}\frac{b_{n+1}}{b_n}=\ell$$, then $$\lim_{n\to\infty}\sqrt[n]{b_{n}}=\ell$$.

In particular, if both limits exist, they have to be equal, so the counter-example the OP is considering does not exist.

Proof: Let $$\epsilon>0$$. Then there exists $$N>0$$ such that, for any $$n\geqslant N$$, we have

$$\ell-\frac{\epsilon}{2}<\frac{b_{n+1}}{b_n}<\ell+\frac{\epsilon}{2}$$

so $$b_n\left(\ell-\frac{\epsilon}{2}\right). By induction, we have

$$b_N\left(\ell-\frac{\epsilon}{2}\right)^{n-N}

so

$$\sqrt[n]{b_N\left(\ell-\frac{\epsilon}{2}\right)^{n-N}}<\sqrt[n]{b_{n}}<\sqrt[n]{b_N\left(\ell+\frac{\epsilon}{2}\right)^{n-N}}$$

But $$\lim_{n\to\infty} \sqrt[n]{b_N\left(\ell-\frac{\epsilon}{2}\right)^{n-N}}=\ell-\frac{\epsilon}{2}$$ and $$\lim_{n\to\infty} \sqrt[n]{b_N\left(\ell+\frac{\epsilon}{2}\right)^{n-N}}=\ell+\frac{\epsilon}{2}$$. This means that for $$n$$ large enough, we have $$\ell-{\epsilon} <\sqrt[n]{b_n}<\ell+\epsilon$$, that is $$\lim_{n\to\infty}\sqrt[n]{b_n}=\ell$$. $$\square$$

Here is an example showing that the Root Test is strictly stronger than the Ratio Test, that is there are situations where we can conclude with the Root Test but not with the Ratio Test:

Consider $$a_n=\left\{ \begin{array}{cc} \frac{1}{3^n}, & n \textrm{ even} \\ \frac{4}{3^n}, & n \textrm{ odd} \end{array} \right.$$, so $$\frac{a_{n+1}}{a_n}=\left\{ \begin{array}{cc} \frac{4}{3}, & n \textrm{ even} \\ \frac{1}{12}, & n \textrm{ odd} \end{array} \right.$$. Therefore, $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$ does not exist and the Ratio Test is inconclusive.

On the other hand, we have $$\lim_{n\to\infty}\sqrt[n]{a_n}=\frac{1}{3}<1$$, so $$\sum_n a_n$$ is convergent by the Root Test.

Another one of my favorite examples is $$a_n=d_nx^n$$ where $$d_n$$ is the number of factors of $$n$$ (for $$n\ge 1$$). The limit of $$\frac{|a_{n+1}|}{|a_n|}$$ does not exist, but the limit of $$\sqrt[n]{|a_n|}$$ is $$|x|$$, so the power series $$\sum_{n=1}^\infty d_nx^n$$ conveges if and only if $$|x|<1$$.

Let $$(b_n)$$ be a sequence such that $$b_n>0$$ for every $$n$$. If $$\lim_{n\to\infty}\frac{b_{n+1}}{b_n}=\infty$$, then $$\lim_{n\to\infty}\sqrt[n]{b_n}=\infty$$.
is true. Let $$M>0$$. There exists $$N$$ such that, for every $$n\ge N$$, we have $$\frac{b_{n+1}}{b_n}\ge 2M$$. So $$b_{n+1}\ge 2M b_n$$. By induction, we have $$b_n\ge (2M)^{n-N}b_N$$ for $$n\ge 1$$. This means that
$$\sqrt[n]{b_n} \ge 2M \sqrt[n]{C}$$
where $$C=\frac{b_N}{2^N}$$. Since $$\lim_{n\to\infty}\sqrt[n]{C}=1$$, then there exists $$N_2$$ such that if $$N\ge N_2$$, then $$\sqrt[n]{C}>\frac{1}{2}$$. Therefore, for $$n\ge Max(N,N_2)$$, we have $$\sqrt[n]{b_n}>M$$. By definition, this means that $$\lim_{n\to\infty}\sqrt{b_n}=\infty$$.
• it is true even if $l$ is $\infty$? Commented May 8, 2023 at 13:58
$$\lim_{n\to\infty}(\log a_{n+1}-\log a_n)=\log L\implies\lim_{n\to\infty}\frac{\log a_n}{n}=\log L$$ by Stolz–Cesàro theorem (explaining the comment by @lhf).