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If $\displaystyle\lim_{n\to\infty} a_n$ does not exist or if $\displaystyle\lim_{n\to\infty}a_n\neq0$, then the series $\displaystyle\sum_{n=1}^{\infty} a_n$ is divergent.

From Stewart's Early Transcendentals Calculus.


I would think that if the the limit $\displaystyle\lim_{n\to\infty} a_n$ does not exist, then it would not equal zero. However, the inclusion that the limit also cannot be zero seems to tell me that I am wrong. In general, if some $a$ is undefined, non-existent, or not even a number, is it good notation to say that $a\neq0$?

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gz839918
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    The word between the two conditions is "or". You seem to be interpreting it as "and". – Joey Zou Jul 16 '16 at 22:20
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    $\lim_{n \to \infty} a_n \neq 0$ means that the limit exists and is not equal to zero. If the limit doesn't exist, the statement is meaningless; it's neither true nor false. – Qiaochu Yuan Jul 16 '16 at 22:23
  • @JoeyZou I think you're misunderstanding the question (or maybe I am). I believe the question is if it would be enough to simply say "If $\lim_{n\to\infty} a_n \neq 0$ then..." and simply leave out the part "If $\lim_{n\to\infty} a_n$ does not exist...". Essentially, if $\lim_{n\to\infty} a_n \neq 0$ covers the case where the limit does not exist. – Eff Jul 16 '16 at 22:23
  • @Eff The case when $(a_n)$ does not exist should not be considered as covered by the assertion that $\lim\limits_{n\to\infty}a_n\ne0$ since this assertion means 1. that the limit exists and 2. that it is not zero. – Did Jul 16 '16 at 22:52
  • @Did I understand that :-). I just explained the meaning of the question (at least as I understand the question). – Eff Jul 16 '16 at 23:01
  • @Eff The trouble is that the sentence "Essentially, if limn→∞an≠0 covers the case where the limit does not exist" in your comment is squarely wrong. – Did Jul 16 '16 at 23:02
  • @Did Exactly. I rephrased the question, I did not answer it. I agree that $\lim_{n\to\infty} a_n \neq 0$ does not cover the case where the limit does not exist, but I didn't want to answer the question in the comments. – Eff Jul 16 '16 at 23:05
  • @QiaochuYuan why would lim$_{n \rightarrow \infty} \neq 0$ not be true or false if the limit did not exist? That statement is true if the limit $is$ $not$ $zero$. – Prince M Jul 17 '16 at 00:05
  • @QiaochuYuan Tell me is this statement true, "If $n$ is both even and odd, then $n$ = 17". – Prince M Jul 17 '16 at 00:08
  • @Prince: that statement is vacuously true. It is not an example of the phenomenon I'm describing. An example is "brown squared equals $17$." This statement is meaningless, and not true or false, because I have not specified what it means to square a color. – Qiaochu Yuan Jul 17 '16 at 00:27
  • @PrinceM However, this is different from what I have asked. I have some $x$, and I want to know whether or not saying that $x\neq0$ also implicates that $x$ has the possibility of being undefined or non-existent. No number can be both even and odd, so your statement can be written, "if FALSE, then n = 17", which really cannot be said without predicate logic, because we are unsure of what $n$ equals. – gz839918 Jul 17 '16 at 00:28

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Firstly, to answer your question, if the limit $\lim\limits_{n\to\infty}a_n$ does not exist, it makes no sense to say that $\lim\limits_{n\to\infty}a_n=a$, where $a$ does not exist. We would say that the sequence $\{a_n\}_{n=1}^\infty$ is divergent. In short, no, it is not good notation to say that $a\neq 0$, in any of the cases you stated.

Secondly, the theorem is stating two separate conditions for the series $\displaystyle\sum_{n=1}^{\infty} a_n$ to be divergent . It could be just as easily (and perhaps less confusingly) have been stated as two separate theorems:

If $\lim\limits_{n\to\infty}a_n$ does not exist, then $\displaystyle\sum_{n=1}^{\infty} a_n$ is divergent.

and

If $\lim\limits _{n\to\infty}a_n\neq0$, then $\displaystyle\sum_{n=1}^{\infty} a_n$ is divergent.

These are two different situations, with no overlap. I think this is where the confusion arises.

Aweygan
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  • That is to say, $a \neq 0$ means that $a$ necessarily exists. (?) – gz839918 Jul 16 '16 at 22:58
  • Lim of sin(x) as x --> infinity. It does not equal 0 and it does not exist. @RobertZhang – Prince M Jul 16 '16 at 23:49
  • Furthermore, saying $\lim\limits_{n\to\infty}a_n=a$ (or a continuous analogue) does not imply equality in the traditional sense. It is simply a notation used to mean that the terms of the sequence (or values of the function) approach that value as the index gets larger. – Aweygan Jul 17 '16 at 03:11
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Suppose I wrote something meaningless and terrible such as

$\sqrt{\mathbb R} + 27 \ne 13 \implies $

$\sqrt{\mathbb R} \ne -14 \implies $

$\mathbb R \ne 196$

Your assumption ought to be that I clearly don't have the foggiest idea what I'm talking about and I am clearly ... lost.

Then supposed I argued "But every statement is true! $\sqrt{\mathbb R}$ is a meaningless expression so it isn't an number and so we can not add 27 to it and we wouldn't get 13 if we did that! So it doesn't equal 13 because it doesn't mean anything."

Well, technically I may be right but I should be banned from teaching for lack of clarification.

$\lim a_n \ne 0$ implies very strongly that $\lim a_n$ is a well-defined value which, although it doesn't equal 0, does equal something else.

Maybe I can quibble $\lim a_n $ diverging $\implies \lim a_n \ne 0$ but no-one can claim that isn't misleading.

At any rate "If $a_n$ diverges then $\sum a_n$ diverges, and if $\lim a_n \ne 0$ then $\sum a_n$ diverges" $\iff$ "If $a_n$ diverges or $\lim a_n \ne 0$ then $\sum a_n$ diverges" $\iff$ "If $\lim a_n \ne 0$ or $a_n$ diverges then $\sum a_n$ diverges" are all equivalent and true statements. Even if a nitpicker might quibble they are redundant. So what? Redundancy is perfectly acceptable.

Especially if it aids clarity.

fleablood
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