1

I would like to find the contrapositive of: If $a_n > 0$ and $\sum a_n$ converges, then $\sum 1/a_n$ diverges.

My first idea is to rewrite it symbolically: Let $a_n$ be a sequence of real numbers. $\forall i \in \mathbb{N}, a_i > 0 ~ \land \sum a_n$ converges $\implies \sum \frac{1}{a_i}$ diverges.

Then we take the contrapositive: $\exists i \in \mathbb{N}, \sum \frac{1}{a_i}$ converges $\implies a_i \le 0 ~ \lor \sum a_n$ diverges.

However, I think this may be incorrect. A more reasonable contrapositive could be:

$\exists i \in \mathbb{N}, a_i \le 0, \sum \frac{1}{a_i}$ converges $\implies \sum a_n$ diverges.

However, I do not know which is right, or more importantly, why a particular one is right (maybe neither are?)

polite proofs
  • 825
  • 2
    The contrapositive of $\phi \Rightarrow \psi$ is $\lnot \psi \Rightarrow \lnot\phi$. Your example is complicated (1) by the ambiguous use of $n$ in $a_n > 0$ (does that mean $\forall n(a_n > 0)$, or what?) and (2) by the possibile ambiguity in the meaning of "diverge" (does $(-1)^n$ diverge?). – Rob Arthan Jan 13 '21 at 00:41
  • 1
    The easiest way to help you is to recommend that you add parentheses in your rewriting. How far into the formula does $\forall i\in\Bbb N$ extend? Is the logical form $(P\land Q)\implies R$ or $P\land(Q\implies R)$? After you clarify those points, the contrapositive should be easier to determine. – Greg Martin Jan 13 '21 at 00:44
  • 1
    @GregMartin I do think that my second one is correct after parentheses. But I do wonder, why isn't the "and" a logical AND? – polite proofs Jan 13 '21 at 00:48
  • 2
    @RobArthan I couldn't really tell you. I am trying to take theorems, and take their contrapositive. – polite proofs Jan 13 '21 at 00:50
  • 1
    I don't understand your question: "and" is definitely a logical AND. (I should point out that neither of your proposed answers is correct.) – Greg Martin Jan 13 '21 at 00:51
  • 2
    @GregMartin Well, because sometimes "and" is used to list numbers. For example, consider: "If 1,2 and 5 are even, then Q." The "and" is not a logical and, it is instead a way to list numbers. – polite proofs Jan 13 '21 at 00:52
  • 1
    May be helpful to re-write it as $\displaystyle \bigg ( \forall \epsilon > 0 ,\exists N, \forall n \ge N, \displaystyle \sum_{j = n}^\infty a_j < \epsilon \bigg ) \wedge \bigg (\forall n, a_n > 0 \bigg )\implies \bigg ( \exists \epsilon > 0, \forall N, \exists n \ge N , \sum_{j=n}^\infty \frac{1}{a_j} \ge \epsilon \bigg )$ – oliverjones Jan 13 '21 at 00:53
  • 1
    @oliverjones Thanks, but does this mean I have to write out the full definitions like that and there is no easy way to take the contrapositive? – polite proofs Jan 13 '21 at 00:56
  • 1
    @oliverjones: does the sequence $a_n = -n$ diverge? I think the OP needs to say more about where the problem comes from. – Rob Arthan Jan 13 '21 at 00:57
  • 1
    @RobArthan The statement in the OP is verbatim from Apostol's Calculus. I wanted to take the contrapositive of it for practice. – polite proofs Jan 13 '21 at 00:57
  • 1
    But what is Apostol's definition of "diverge"? (I think we can safely guess that $a_n > 0$ is intended to mean $\forall n(a_n > 0)$.) – Rob Arthan Jan 13 '21 at 01:00
  • 1
    @RobArthan I think so too, that's why I wrote it like that in my post – polite proofs Jan 13 '21 at 01:01

1 Answers1

1

After the long exchange of comments, I think the best answer we can give is that the contrapositive of:

If, for every $n$, $a_n > 0$, and, if $\sum_n a_n$ converges, then $\sum_n 1/a_n$ diverges: i.e., $$(\forall n (a_n > 0)) \land C(a)) \Rightarrow D(n \mapsto 1/a_n)$$

is:

if $\sum_n 1/a_n$ does not diverge, then, either, for some $n$, $a_n \le 0$ or $\sum_n a_n$ does not converge: i.e., $$\lnot D(n \mapsto 1/a_n) \Rightarrow (\exists n (a_n \le 0)) \lor \lnot C(a)$$

where $C(a)$ and $D(a)$ stand for whatever Apostol means by $\sum_n a_n$ converges or $\sum_n a_n$ diverges, respectively.

Rob Arthan
  • 48,577
  • I see. Thank you. However, just to be sure, the “either” doesn't mean exclusive OR, correct (as you didn't write so symbolically)? It could be the case that it both diverges and that there's an $a_i \le 0.$ – polite proofs Jan 13 '21 at 01:55
  • 1
    Yes, that's an inclusive OR. – Rob Arthan Jan 13 '21 at 02:46