Question

I have another lovely convergence question: If one of the absolute convergence tests always results in convergence, divergence, or inconclusive (at which point you move on to another absolute convergence test), then how do you end up testing for conditional convergence?

Answer 1

If that makes any sense at all

Answer 2

For example, if you tried the ratio test on a series and it diverges, then doesn't that mean that the series diverges? I dont understand how the AST would ever come into play

Answer 3

If \(\sum |a_n|\) is convergent, then the series converges absolutely. If \(\sum |a_n|\) diverges but \(\sum a_n\) converges, the series is conditionally convergent.

Answer 4

And you would prove that the series itself converges through the AST?

Answer 5

That is one test to show convergence, yeah.

Answer 6

Ok. So then in order to prove that a series diverges, you have to show that it does not converge conditionally or absolutely?

Answer 7

So like proving that the AST (the only test we have been given thus far that we can use if absolute convergence doesn't work) and the ratio test dont work

Answer 8

To show that the series is divergent, you would have to show that \(\sum a_n\) diverges. That would imply that \(\sum|a_n|\) also diverges.

Answer 9

Either test you mentioned could do the job.

Answer 10

What tests prove general divergence?

Answer 11

Oh I guess the AST can't prove divergence. If you get a limit > 1 in, for example, the ratio test applied to \(\sum a_n\) that would prove that the series is divergent.

Answer 12

Does that prove that the series is divergent or just that it does not absolutely converge

Answer 13

It seems to me like the ratio test (and comparison tests and integral test) only prove absolute convergence

Answer 14

Its hard to put my confusion into words. I have a problem that I guess would clear up a lot

Answer 15

The ratio test, integral test, and comparison test can all show divergence. If you apply the integral test or comparison test to \(\sum a_n\) and it converges, the series is at least conditionally convergent. If the sum diverges, the series is divergent. If you apply the integral or comparison test to \(\sum |a_n|\) and it converges, then the series is absolutely convergent, and \(\sum a_n\) is also convergent. If \(\sum a_n\) is divergent, then the series may or may not be conditionally convergent.

If the ratio test has a limit < 1, the series is absolutely convergent. If it gives a limit > 1, the series is divergent. limit = 1 is indeterminate.

Answer 16

Maybe this can help
http://tutorial.math.lamar.edu/Classes/CalcII/AbsoluteConvergence.aspx

Answer 17

I see. Can you help me with my problem? I think that the concept would make more sense to me if I could figure it out

Answer 18

I can try, but I am a bit rusty :P

Answer 19

Lol ok. I appreciate it. I am supposed to find the interval of convergence (including conditional and absolute) for \[\sum_{n=0}^{\infty}\frac{ nx^n }{ 4^n(n^2+1) }\]

Answer 20

ok i will probably have to read up for a sec, so give me a moment please

Answer 21

The ratio test came out and I got (-4,4), but when I tested -4 with the limit comparison test I got that it diverged, so I was confused about whether or not I am allowed to go onto the AST

Answer 22

If the limit comparison test showed divergence, then the series is divergent.

Answer 23

The answer, though, is that the series is conditionally convergent at -4

Answer 24

Heres the solution if that helps http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/523/exercises/44a/

Answer 25

It says the series is absolutely convergent on the interval \([-4,4)\). I'm not sure why it's an open interval, but I understand the rest of the answer as it is stated in the link.

Answer 26

It says for part c that the series converges conditionally at x=-4

Answer 27

It doesn't make any sense. It seems like my book contradicts itself

Answer 28

Yeah I can't really make sense of it. It may just be my lack of practice lately. @amistre64 can help maybe

Answer 29

I appreciate it!

Answer 30

ive got too much lag to do much of anything at the moment

Answer 31

\[\sum_{n=0}^{\infty}\frac{ n(-1)^n(4)^n }{ 4^n(n^2+1) }\]
\[\sum_{n=0}^{\infty}\frac{ n(-1)^n}{ (n^2+1) }\]
\[\sum_{n=0}^{\infty}\frac{ n(-1)^n}{ (n^2+1) }\implies\sum_{n=0}^{\infty}\frac{ n}{ (n^2+1) }\]
\[(1)~~\lim_{n\to \infty}b_n=0~and;\\
(2)~~\{b_n\}~is~decreasing~sequence\\\]
\[then~\sum a_n~converges\]

Answer 32

http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

Answer 33

What happened to testing for absolute convergence? You are always supposed to do that first

Answer 34

good luck

Answer 35

Wat

Answer 36

Noooooo

Answer 37

@amistre64 you didn't answer my question

Answer 38

you see how picking x=4, means it diverges, tahts why that point is not included

Answer 39

I know. My confusion is why they used the alternating series test after they proved that the series diverged through an absolute convergence test

Answer 40

after they found the interval of convergence?

Answer 41

Yes. When they were testing the endpoints

Answer 42

Well they skipped proving absolute convergence anyways

Answer 43

well amistre there is using aBS convergence test there

Answer 44

He used the alternating series test, which I dont think he is allowed to do until he proves absolute convergence doesn't happen

Answer 45

http://prntscr.com/6vtqg3

Answer 46

no he's showing the Abs value of that series is convergent as long as x<4 on the positive bound

Answer 47

Either way that doesn't help

Answer 48

okay i see

Answer 49

conditionally convergent at -4, means its ABS is divergent and AST is convergent right?

Answer 50

yes

Answer 51

But I thought that if an absolute value series diverged then that shows that the series diverges

Answer 52

Oh no, that is not necessarily true

Answer 53

you can even without any analysis intuitively tell that Alternating series terms would have a smaller ABS value in the end than the ABS value of all positive sums