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The ratio test says that, for $a_k\neq 0$, if $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=L$$ exists, then if $0\leq L <1$, then $\sum_k a_k$ converges. If $L>1$, it diverges.

The notes I'm reading say that it's inadmissible to use the ratio test to test for convergence of a geometric series. I can't see why this should be the case.

Say we have some geometric series $\sum_kar^k$. Then $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{\left|ar^{k+1}\right|}{\left|ar^k\right|}=|r|.$$ So the ratio test tells us that the geometric series converges for $|r|<1$, and diverges for $|r|>1$, which is exactly what we get by using the formula $$\sum_{k=1}^n ar^k=a\left(\frac{1-r^{n+1}}{1-r}\right).$$

What is an example that demonstrates why the ratio test is inadmissible for a geometric series?

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    $\begingroup$ How do you prove the ratio test? Probably by invoking the theorem that the geometric series converges. $\endgroup$ Sep 6, 2017 at 17:29
  • $\begingroup$ Makes sense, thanks $\endgroup$ Sep 6, 2017 at 17:29
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    $\begingroup$ Aside from the possibly circular logic in proving the ratio test, it certainly applies to geometric series and gives the correct result as you showed. $\endgroup$
    – user169852
    Sep 6, 2017 at 17:55

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The usual proof of the ratio test is to compare the series to a geometric series. If $$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \alpha < 1,$$ then we have $$ |a_{n+1}| < |a_n| \alpha $$ for all sufficiently large $n$. It then follows from an induction argument that $$ |a_{n+k}| < |a_n| \alpha^k $$ for $n$ sufficiently large, which means that $$ \sum_{j=1}^{\infty} |a_j| = \sum_{j=1}^{n-1} |a_j| + \sum_{j=n}^{\infty} |a_j| \le \sum_{j=1}^{n-1} |a_j| + \sum_{k=0}^{\infty} |a_n| \alpha^k.$$ The first series has only finitely many terms and is therefore finite, and the second series is geometric and therefore converges by some other argument. From this, it follows that if $$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1, \qquad\text{then}\qquad \sum_{j=1}^{\infty} a_j $$ converges (by the limit comparison test, for example). The other inequality in the ratio test can be argued by noting that the general term does not go to zero, and the uncertainty at 1 can be argued by considering (for example) the harmonic and alternating harmonic series.

The key point is that we prove that the ratio test implies the convergence of series by comparison to a convergent geometric series. Since the proof of the ratio test relies on this convergence, it is circular to argue that a geometric series converges by the ratio test (unless, of course, you have another proof for the ratio test that doesn't use the convergence of geometric series).

While one could use the ratio test to establish the convergence of a geometric series (there is nothing stopping us!), it is typically poor style to rely on circular arguments as it can (potentially) lead to overlooking important hypotheses or exceptional cases. This is particularly important in a pedagogical setting, when students may not be entirely cognizant of the line of reasoning that lead up to a result (it is hard to keep track of all of the lemmata, theorems, and proofs that lead up to a result if it is the first time that you have had to deal with them).

Moreover, I don't see why one would want to use the ratio test to show that a geometric series converges. Basically no computation is needed to show that a geometric series converges, while a couple of computational steps are needed in order to invoke the ratio test.

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  • $\begingroup$ Or, you've already proven the ratio test using the convergence of geometric series, and since the former subsumes the latter there's little point in retaining the latter. $\endgroup$
    – user14972
    Sep 7, 2017 at 8:55
  • $\begingroup$ @Hurkyl Note that I have not said that it is "inadmissible" to use the ratio test to show that a geometric series converges, only that it is "circular". $\endgroup$
    – Xander Henderson
    Sep 7, 2017 at 12:53
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We can't know for sure (unless context from the notes you've omitted says something on the topic), but the various comments appearing in this topic suggest an explanation.

Two major activities mathematicians (and people who use mathematics) engage in are:

  • Using mathematical tools to do calculations and prove things
  • Designing and validating the mathematical tools used in the above bullet

The ratio test is an example of such a mathematical tool, and is perfectly applicable to geometric series. It may even be the preferable tool for deciding when a geometric series converges, simply to cut down on the number of things one needs to remember.

However, if you happen to be in the process of validating the ratio test, it would not be valid to use the ratio test to justify any of the facts you need — such as when geometric series converge — in its validation. You would have to first derive the facts about geometric series in some other fashion.

As for your notes, the most likely explanations are:

  • You are reading notes about how to validate the ratio test
  • You have misunderstood the notes
    • Possibly due to the author failing to actually convey his meaning
  • The author of these notes was confused

To elaborate on that last point, people often get stuck in the "tool building" mindset. A fair amount of mathematics education involves validating the tools one is already familiar with, which requires temporarily suspending the use of those tools so that we can see how they can be built from more basic tools — and sometimes people learn the wrong lesson and think that's something you always have to do.

So, when facts about convergence of geometric series in the proof of the ratio test, this sometimes leads to people mistakenly thinking that any time the topic of convergence of geometric series comes up, one must suspend their use of the ratio test and argue from more basic tools.

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  • $\begingroup$ Here are the notes (look at the second bullet point on page 46) courses.maths.ox.ac.uk/node/view_material/1087 $\endgroup$ Sep 7, 2017 at 7:57
  • $\begingroup$ and then point B on the problem sheet (page 2): "Why is it not admissible to establish the convergence or divergence of a geometric series by using the Ratio Test?" courses.maths.ox.ac.uk/node/view_material/347 $\endgroup$ Sep 7, 2017 at 7:58
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    $\begingroup$ @man_in_green_shirt: Unless that road sign in the margin has some relevant significance, the notes are simply wrong as written, and I think my answer is accurate. I think the best odds are on the author wanting to talk about the specific setting of how one would prove whether or not a geometric series converges if this fact was not already known, but failing to properly frame the remark. $\endgroup$
    – user14972
    Sep 7, 2017 at 8:53
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The ratio test is not inadmissible for geometric series. Its hypotheses do not exclude geometric series, therefore it applies, and its proof must support this.

One common proof structure would be:

Theorem A: Geometric series converges. Proof: direct argument.

Theorem B: Ratio test with usual hypotheses. Proof: show that this is implied by theorem A as in the answer by Xander Henderson.

Of course, the proof of Theorem A cannot use Theorem B, otherwise we have a circular argument. Undoubtedly this is what your notes are trying to say.

However, once we have a valid proof of Theorem B, it certainly applies to geometric series:

Theorem C: Geometric series converges. Proof: theorem B.

This proof of Theorem C may seem absurdly indirect: why wouldn't we just cite Theorem A? Well, consider:

Theorem D: some other theorem whose hypotheses imply those of the ratio test. Proof: Theorem B.

It would be very annoying, and more importantly unnecessary, to instead write:

Theorem D: some other theorem whose hypotheses imply those of the ratio test. Proof: If the series is geometric, then see Theorem A. Otherwise Theorem B applies.

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Once we have a valid proof of the ratio test, then it can be applied to any series if it satisfies the hypotheses, including geometric series. It does not matter how the ratio test was proven. This is in no way circular reasoning even if the proof used results about geometric series. The idea of circular reasoning is a logical fallacy, but that is not what is being done here. We are not assuming what we are trying to prove. In fact, if we are allowed to assume statement $A$ because it is given to us, then the proof of $A$ is immediate from our given assumption. This is not circular reasoning because we are explicitly allowed to use statement $A$ which is somehow given to us to use.

The OP gave a link to course notes that state 'We cannot prove that a geometric series $\sum r^k$ is convergent/divergent by applying the Ratio Test. Why not?" That's a convincing argument, isn't it? Despite the hypotheses for "D'Alembert's Ratio Test" are satisfied by any positive geometric series.

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  • $\begingroup$ It's correct but still circular, like computing $\lim_{x \to 0} \frac{e^x-1}{x}$ using L'Hospital's rule. Ultimately if you expanded out the proof of the ratio test, you would find that in the body of that proof is the result you already want to prove being asserted without proof...unless you have some other proof of the ratio test. $\endgroup$
    – Ian
    Sep 6, 2017 at 18:32
  • $\begingroup$ @Ian We can easily fix the circularity by embedding a direct proof of the convergence of geometric series in the proof of the ratio test. If we assume this is done, then there's no logical problem applying the ratio test to geometric series, even if it is unnecessary to do so. $\endgroup$
    – user169852
    Sep 6, 2017 at 19:22
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    $\begingroup$ @Bungo It's true, but then you are effectively saying "I have a black box containing the desired proof. Showing you this black box without opening it constitutes a proof." It's logically sound but socially flawed, and proof is a social activity. $\endgroup$
    – Ian
    Sep 6, 2017 at 19:30
  • $\begingroup$ @Ian I would argue that any proof of the ratio test that uses without proof the fact that a geometric series converges, is inadequate. Otherwise, every time we want to prove a new theorem whose hypotheses imply those of the ratio test, we would have to bifurcate into two cases: "if $x_n$ is not geometric, then the ratio test applies; otherwise, it converges by the geometric series convergence theorem". If the ratio test hypotheses do not exclude the geometric series, then the proof must include or cite a proof of this convergence (or not depend on this convergence). $\endgroup$
    – user169852
    Sep 6, 2017 at 20:04
  • $\begingroup$ @Bungo In my analogy, you're proving the geometric series converges, you present a black box that is a proof of the ratio test, and you don't open the proof of the ratio test, so you don't get to see the details of the proof that the geometric series converges, but you "promise to the listener" that it is there anyway. $\endgroup$
    – Ian
    Sep 6, 2017 at 20:10

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