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Rudin's Principles of Mathematical Analysis phrases the ratio test (Thm. 3.34) as follows:

"The series $\sum a_n$

  1. converges if $\limsup_{n \to \infty} |a_{n+1}/a_n| < 1$,

  2. diverges if $|a_{n+1}/a_n| \geq 1$ for $n \geq n_0$, where $n_0$ is some fixed integer.

  3. If $$\liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \leq 1 \leq \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|,$$ the test gives no information."

My assumption was that for every series, exactly one of the conditions in (1)-(3) must hold.
From the phrasing this wasn't immediately obvious, and after thinking about it a bit I found that it isn't actually the case.
For example the series $1 + 1 + 1 + \cdots$ satisfies both (2) and (3).
I think we can fix this overlap by rephrasing the condition of (2) as $\liminf > 1$.
I like this better because it makes the conditions more symmetrical and it's obvious then that exactly one of the three must be satisfied in all cases.

Does this leave the theorem intact?

Edit: I think the change does leave the theorem true and fix the logical overlap, but weakens it because (2) is true as written and my replaced condition for (2) is narrower. Is there a way to keep the theorem true without weakening it but still fix the logical overlap in a simple way?

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  • $\begingroup$ Btw (b) should be "for all $n\ge n_0$". $\endgroup$ Dec 21, 2019 at 15:18

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By changing $\lim\inf\ge 1$ to $\lim \inf>1$ in condition (b), this "leaves the theorem intact" in the sense that it is still true. After all, if a series is divergent when $\lim\inf\ge 1$, then it is clearly divergent if $\lim\inf>1$.

However, as plasticRing noted in his answer, removing the equality condition actually weakens the theorem in the sense that we can no longer prove divergence in some instances. It is always more useful to know for a fact that a series diverges than to have no idea. Your example of $a_n=1$ illustrates this perfectly. By satisfying (b) (without the change), we can say definitively that $1+1+1+\cdots$ is divergent. With the change you've suggested, we can't draw any conclusions from this test.

Remark: It would probably be clearer if phrased "if (a); else if (b); else if (c)."

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  • $\begingroup$ Does the original condition in (b) mean exactly that $\liminf \geq 1$? $\endgroup$
    – Nick A.
    Dec 19, 2019 at 22:51
  • $\begingroup$ @NickA.: No; consider, e.g., the case where $a_{n+1}/a_n=1-1/n$. $\endgroup$
    – Micah
    Dec 19, 2019 at 23:14
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You lose information for cases in which $\liminf =1$ in the special form of the original (b). What seems better is to add to (c) 'if (b) is not satisfied and '.

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I've always hated that traditional "If [inequality] the ratio test gives no information". Depending on exactly what it means for the ratio test to "give information", if that sentence is part of the ratio test it's self-contradictory. And in any case, if in a certain situation a theorem gives no information it's just silly to state as part of the theorem that it gives no information in that case; in fact it doesn't give any information, whether it says so or not.

What's meant is just "If [inequality] then the series may converge or diverge". That's clearer, or at least certainly no less clear, and this problem goes away.

He must have felt the same way, because the problem doesn't arise in my copy of the book. There's no part (c) there; instead we see (a) and (b), and then after the proof of the theorem it says

"Note: The knowledge that $\lim\dots = 1$ gives no information about the convergence of $\sum a_n$. The series $\sum 1/n$ and $\sum 1/n^2$ illustrate this."

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