In my analysis homework, I have the following two True/False questions about limits:

(c) if

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$$

Then $(a_n)$ converges

(d) if

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$$

Then $(a_n)$ diverges

The thing is: I'm not sure how to interpret these questions. In the answer key, both are given as False; for (c) they give the example where $a_n = n$, and for (d) they give $a_n=\frac{1}{n}$.

Here's where I get confused:

  1. For both (c) and (d), how do I interpret the question exactly? It seems that they give counterexamples where it couldn't be the case that $(a_n)$ converges / diverges.

  2. The way I interpret it now is that the limit of the absolute value of $\frac{a_{n+1}}{a_n}$ as n tends to infinity is 1. So if we have $a_n =1$, how is this not the case for the sequence $(a_n)$?

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    $\begingroup$ Question c) means: Does it hold that any sequence $(a_n) $ satisfying $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$ converge? Can you translate question d)? $\endgroup$ – mfl Oct 6 '17 at 15:33
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    $\begingroup$ The counter-examples show that "limit=$1$" does not give any information whether the sequence converges or diverges. So, both implications are false. Neither must the sequence converge, nor must it diverge. $c)$ claims it must converge , $d)$ claims it must diverge $\endgroup$ – Peter Oct 6 '17 at 15:34
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    $\begingroup$ @Peter You're right.......It reminds me of a Ratio test in Calculus to determine if a given series will converge or diverge. Evaluating $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, the $a_n$ is converging.....if $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| > 1$, then $a_n$ is diverging......If $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, the test fails $\endgroup$ – Palautot Ka Oct 6 '17 at 15:50
  • $\begingroup$ @mfl: this is the source of my confusion. I only started with limits last week, and as I understand it, any sequence that has a limit converges. So, as I understand it now, If we take $a_n=1$, as in the answer key, then as n tends to infinity $(a_n)$ should converge to 1. $\endgroup$ – JHG90 Oct 6 '17 at 17:28
  • $\begingroup$ @Peter: could you expand on 'gives no information'? $\endgroup$ – JHG90 Oct 6 '17 at 17:29

There's a default convention behind such questions. Informally speaking: any such statement with some object in it is understood to be true if it is always true, i.e. if it is true for any such object.

For example, the statement

(c) If $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$, then $(a_n)$ converges,

actually is meant to say

(c) For any sequence $(a_n)$, if $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$, then $(a_n)$ converges,

and that's what you need to classify as true or false.

  • The statement is true if it is true for all such sequences $(a_n)$;
  • The statement is false if at least one counterexample exists (regardless of the fact that there may be other examples that satisfy it).

The answer key for this question gives you such a counterexample: the sequence $a_n=n$ satisfies the "if" part since $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty} \frac{n+1}{n}=1$, and yet $\displaystyle \lim_{n\to\infty}a_n$ doesn't exists since these numbers tend to infinity.

Question d) should be interpreted similarly.

And then if you put them together, then as @Peter said, we see that the fact $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$ gives us no information about the behavior of the sequence $(a_n)$: some such sequences, like the counterexample for c), diverge; and some such sequences, like the counterexample for d), converge.


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