If $\lim a_n$ does not exist, then does this mean that $\lim a_n\neq 0$?

Question

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 $a\neq0$?

Answer 1

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.

Answer 2

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.