Ratio test and root test

Question

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.

Answer 1

As mentioned in the comments:

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 $b_n\left(\ell-\frac{\epsilon}{2}\right)<b_{n+1}<b_n\left(\ell+\frac{\epsilon}{2}\right)$. By induction, we have

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

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$. $\square$

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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 $|x|<1$.

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To answer a comment:

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 $\lim_{n\to\infty}\sqrt{b_n}=\infty$.

Answer 2

$$\lim_{n\to\infty}(\log a_{n+1}-\log a_n)=\log L\implies\lim_{n\to\infty}\frac{\log a_n}{n}=\log L$$ by Stolz–Cesàro theorem (explaining the comment by @lhf).