The Second Ratio Test is a relatively new convergence test, which we can give as

Let $\sum_n a_n$ be a series with positive terms. For $k \in \{0,1\}$ we write $$ L_k = \limsup_{n \to \infty} a_{2n+k}/a_n , \quad l_k = \liminf_{n \to \infty} a_{2n+k}/a_n $$

  • If $ \max\{ L_0,L_1 \} < 1/2 ,$ $\sum_n a_n$ converges.

  • If $ \min\{ l_0,l_1 \} > 1/2 , $ $\sum_n a_n$ diverges. Otherwise, the test is inconclusive.

(The second condition can probably be replaced by $a_{2n+k}/a_n \geq 1$ for all sufficiently large $n$ if we want to be as strong as possible.)

We recall that a convergence test A is stronger than one B if A resolves the convergence/divergence of any series that B resolves. (One may wish to separate the convergence test from the divergence test for greater specificity.) For example, we all know that the Root Test is stronger than the Ratio Test. A natural question is thus if the Second Ratio Test has a similar relationship to the Root Test, or other more venerable tests. Currently known:

  • The Second Ratio Test resolves the convergence/divergence of the $p$-series $\sum_n n^{-p}$, for $p \neq 1$ anyway. The Root Test does not.
  • The Second Ratio Test is stronger than the Ratio Test and Raabe's Test, at least on the convergence side.

These are proved in this paper cited in the Wikipedia article. Also:

  • The Second Ratio Test requires $a_n \neq 0$, whereas the Root Test avoids this. But this still raises the question of whether one is stronger than the other in their common domain of validity.
  • The Second Ratio Test is not as strong as the Condensation Test where both apply: it does not resolve $n^{-1} (\log n)^{-p}$, whereas the Condensation Test does. But the Condensation Test requires a monotonically decreasing sequence of terms. In this case, there is a follow-up paper that discusses more general cases.
  • More interestingly, Bertrand's Test (third in the De Morgan Hierarchy) does resolve $n^{-1} (\log n)^{-p}$, so the Second Ratio Test is not stronger than Bertrand's Test.

(For the definitions of the other tests mentioned above, the linked Wikipedia article has them.)

So this leaves two questions:

  1. If $a_n>0$, is the Second Ratio Test stronger than the Root Test?
  2. Is Bertrand's Test stronger than the Second Ratio Test?

The answer to both questions is no -- we can construct counterexamples using series which are constant over long intervals.

Question 1: consider the series $\sum_n a_n$ such that $a_n = 2^{-3^{k+1}}$ for $3^k \leq n < 3^{k+1}$. This is convergent by the root test, since for any $n$, $|a_n| < 1$, hence taking the appropriate value of $k$, we have $|a_n|^{1/n} \leq |a_n|^{1/3^{k+1}} = \frac{1}{2}$, so $\limsup |a_n|^{1/n} \leq \frac{1}{2}$. The second ratio test is inconclusive, since $\frac{a_{2n}}{a_n} = 1$ when $n = 3^k$, meaning $L_0 \geq 1$.

Question 2: consider the series $\sum_n b_n$ such that $b_n = 4^{-k}$ for $2^k \leq n < 2^{k+1}$. This is convergent by the second ratio test, since $L_0 = L_1 = \frac{1}{4}$, but Bertrand's test is inconclusive, as the term $$\ln n \left(n \left(\frac{b_n}{b_{n+1}} - 1\right) - 1\right)$$ becomes arbitrarily large for $n$ of the form $2^k - 1$, while it is negative for all other $n$.

  • $\begingroup$ Interesting. The second one dashes all hope of further comparison with the de Morgan tests. However, I wonder if there is an "expanded" ("evaporated"?) series of this kind that works in the first question for all $m$th ratio tests, since examples of the present sort will have problems with sufficiently large $m$. $\endgroup$
    – Chappers
    Mar 18 '20 at 22:59
  • 1
    $\begingroup$ assuming I'm interpreting your question right, try $a_n = 2^{-2^{(k+1)^2}}$ for $2^{k^2} \leq n < 2^{(k+1)^2}$; this is still convergent by the root test but every $m$-th ratio test should be inconclusive $\endgroup$
    – user125932
    Mar 18 '20 at 23:08
  • $\begingroup$ Yes, that should do. Cool. $\endgroup$
    – Chappers
    Mar 18 '20 at 23:37

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