Prove that this proposition about series is false

[dunkelheit]

Find a counterexample for the proposition
[MATH]\sum_{n=1}^\infty a_n < \infty \Rightarrow \sum_{n=1}^\infty \frac{|a_n|}{n} < \infty[/MATH]

The counterexample is not hard to find, but I think that is much easier using the contrapositive: I think that the contrapositive is
[MATH]\sum_{n=1}^\infty \frac{|a_n|}{n} = \infty \Rightarrow \sum_{n=1}^\infty a_n = \infty[/MATH]So for the contrapositive I would use [MATH]a_n = \frac{1}{n}[/MATH], hence
[MATH]\sum_{n=1}^\infty \frac{\left|\frac{1}{n}\right|}{n} = \sum_{n=1}^\infty \frac{1}{n^2} < \infty \Rightarrow \sum_{n=1}^\infty \frac{1}{n} = \infty [/MATH]So the contrapositive is false and hence the original proposition is false too.
So my doubts are:
1) is the contrapositive of the initial proposition actually the one I've written before or am I wrong?
Would it be the same if I write the contrapositive as it follows
[MATH]\sum_{n=1}^\infty a_n = \infty \Rightarrow \sum_{n=1}^\infty \frac{|a_n|}{n} = \infty[/MATH]Or not?
2) In general, finding a counterexample for the contropositive is equivalent to finding counterexample for the non contropositive? I think yes because they're logically equivalent, but I'm not sure.
Thanks.

 

[pka]

The counterexample is not hard to find, but I think that is much easier using the contrapositive: I think that the contrapositive is
[MATH]\sum_{n=1}^\infty \frac{|a_n|}{n} = \infty \Rightarrow \sum_{n=1}^\infty a_n = \infty[/MATH]

So post your counter-example so that we can see what you mean. And we can check your work.

 

[Dr.Peterson]

How does your example make the contrapositive false? Its condition is not true.

(1) You have the right contrapositive; your alternative is the inverse, not the contrapositive.

(2) Yes.

 

[dunkelheit]

@pka: the counterexample is done with the sequence [MATH]a_n=\frac{(-1)^n}{\ln (n+1)}[/MATH], we have that for the alternating series test
[MATH]\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln (n+1)} < \infty[/MATH]But for comparison with generalized harmonic series is
[MATH]\sum_{n=1}^{\infty} \frac{\left|\frac{(-1)^n}{\ln (n+1)}\right|}{n} =\sum_{n=1}^{\infty} \frac{1}{n \ln (n+1)} = \infty[/MATH]
@Dr.Peterson: Okay, maybe now I get it: so the contrapositive is well written but my example fails because, using the contrapositive, I have to show an [MATH]a_n[/MATH] such that [MATH]\frac{|a_n|}{n}[/MATH] has divergent series but [MATH]a_n[/MATH] has a convergent series in order to get a counterexample for the contrapositive. Am I right?

Thanks to both of you.

 

[pka]

@pka: the counterexample is done with the sequence [MATH]a_n=\frac{(-1)^n}{\ln (n+1)}[/MATH], we have that for the alternating series test
[MATH]\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln (n+1)} < \infty[/MATH]But for comparison with generalized harmonic series is
[MATH]\sum_{n=1}^{\infty} \frac{\left|\frac{(-1)^n}{\ln (n+1)}\right|}{n} =\sum_{n=1}^{\infty} \frac{1}{n \ln (n+1)} = \infty[/MATH]

I think that you could profit from studding the logic of counterexamples. Counterexample can disprove a proposition but you know that an example cannot prove a proposition.

 

[dunkelheit]

@pka: You're right, I wrote "prove" in the title but then I wrote "find a counterexample" in the text of the exercise, of course I wanted to disprove it (sorry for the time loss).

 

[Dr.Peterson]

@pka: You're right, I wrote "prove" in the title but then I wrote "find a counterexample" in the text of the exercise, of course I wanted to disprove it (sorry for the time loss).

But "prove ... false" is the same as "disprove", for which a counterexample is appropriate. Your title is fine, and I read it as intended.

On the other hand, the counterexample in post #4 is not what you originally used, which briefly confused me.

 

[dunkelheit]

@Dr.Peterson: Oh okay, sorry, I thought it was clear: in #4 I show a counterexample for the original statement, while I used [MATH]\frac{1}{n}[/MATH] as a counterexample for the (badly written) contrapositive statement (so the latter is obviously wrong, because there was a flaw in my understand of how the contrapositive works). I hope that now it is clear, sorry for the bad organization of my answers.
Saying that, is what I've said in #4 correct (if it is understandable )? Thanks again.