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
are nonnegative numbers.
Re: Comparision test for convergence
If for all n,
("for all but a finite n" is sufficient) and
, then if
converges, so does
.
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
to
, it just refers to the size of the series.
Re: Comparision test for convergence
So if you show that the
is greater than or equal to
, does this imply
for
?
Re: Comparision test for convergence
Quote:
Originally Posted by
Duke
So if you show that the

is greater than or equal to

, does this imply

for

?
No it does not. That is backwards.
. Not the other way round.
Re: Comparision test for convergence
ok thanks for that. I am still confused about 2 things. One is that
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
with the corresponding term in
so how does the proof work?