Comparision test for convergence

I was wondering whether one showing series is greater than another is sufficent to show if the former converges, so does the latter. Thats what I took the proof in the link to imply. Cauchy condensation test - Wikipedia, the free encyclopedia. But I've figured that must be wrong. So how does this proof justify convergence simply by noting one series is bigger than the other?

Thanks

Re: Comparision test for convergence

What do you mean exactly by "a series is greater than another". It works if the $\displaystyle a_n$ are nonnegative numbers.

Re: Comparision test for convergence

If for all n, $\displaystyle a_n\le b_n$ ("for all but a finite n" is sufficient) **and** $\displaystyle a_n\ge 0$, then if $\displaystyle \sum b_n$ converges, so does $\displaystyle \sum a_n$.

Re: Comparision test for convergence

I mean, as it shows on the link, the infinite sum. The difficulty I have with the proof is that it doesn't compare $\displaystyle a_n$ to $\displaystyle b_n$, it just refers to the size of the series.

Re: Comparision test for convergence

So if you show that the $\displaystyle \sum b_n$ is greater than or equal to $\displaystyle \sum a_n$ , does this imply $\displaystyle b_n\ge a_n$ for $\displaystyle n\ge N$?

Re: Comparision test for convergence

Quote:

Originally Posted by

**Duke** So if you show that the $\displaystyle \sum b_n$ is greater than or equal to $\displaystyle \sum a_n$ , does this imply $\displaystyle b_n\ge a_n$ for $\displaystyle n\ge N$?

No it does not. That is backwards.

$\displaystyle 0 \leqslant a_n \leqslant b_n \, \Rightarrow \,\sum\limits_n {a_n } \leqslant \sum\limits_n {b_n } $. Not the other way round.

Re: Comparision test for convergence

ok thanks for that. I am still confused about 2 things. One is that $\displaystyle \sum a_n$ does not always exist so how does the implication work then. Second, the proof of the cauchy condensation theorem (on wikipedia) does not compare each term in $\displaystyle a_n$ with the corresponding term in $\displaystyle b_n$ so how does the proof work?