Test the following series for convergence or divergence. Justify the decision.

The series diverges. To show this we use the limit comparison test (more precisely, we use the comment following the proof of the limit comparison test that if then converges implies converges) with

Then we have

Hence, the convergence of would imply the convergence of , but we know diverges; hence, must diverge as well.

Hello sir, and thank you so much for posting these solutions, you are really a life-saver.

But I’m having some problems related to your use of the limit comparison method.

The theorem on apostol says that it convergence of a implies on convergence of b if the quotient equals to 1, and you used it to some extent when the quotient equals 0, what is your logic?

It is easy to extend the theorem no any constant c > 0 but not to 0.

Could you elaborate?

Thx in advance

Hi, I’m using the comment after Theorem 10.9 on page 396 of Apostol. The comment tells us that if

then we can conclude that the convergence of implies the convergence of . The idea of this (without giving a formal proof) is just that the must be smaller than the for all (otherwise the limit couldn’t be going to 0). As you point out, the full theorem doesn’t hold in this case since might converge, but this would not imply the convergence of .

In this case though we are saying that if converged it would imply that converges. Since we know does not converge, this means cannot converge either. Does that make sense?

Oh, it makes sense now (I did not notice the comment on that theorem).

I googled a lot and couldn’t find a formal proof, but as much as “waving hands” go this is really intuitive.

Thx a lot man, I must say I am a big fan of yours (Just started my calculus 3 course and this is helping a LOT) You are a really bright person, hahaha.

Ha, thanks, and no problem. I think you can get to a formal proof of the comment by following Apostol’s proof of the theorem, but leaving off one side of the inequality. So, since we know there is some such that for all we have (just taking in the definition of the limit) and so for all . So, then by Theorem 10.8 we have that converges implies converges. (This still isn’t totally rigorous, but it’s closer.)

Understood your Idea and agree with it, thank you very much once again.

It is just the contrapositive

Why isn’t Rori’s explanation (in the second reply) totally rigorous? I can’t really find any problems with it.

This is a very good question. And RoRi answered it very well. I would like to add that logically this statement:

is logically equivalent to:

and is called the contrapositive.

It can be be proven by contradiction:

Let , and assume for the sake of contradiction that it is not true that . Which means we have B not being true and A being true. But that is a contradiction since A being true implies B being true.

The converse (for equivalence) can be proven similarly.