Question

I had a quick question about series: Which tests prove convergence, and not absolute or conditional convergence?

Answer 1

@rational

Answer 2

Except for alternating series test, almost all other tests require terms to be positive right ?

Answer 3

Umm yeah

Answer 4

I was just thinking that I want to prove convergence, and show that it doesn't converge absolutely

Answer 5

then i think you may use alternating series test to prove "convergence of actual series"

and use any other test to prove "convergence/divergence of absolute value series."

Answer 6

I thought that the alternating series test proved conditional convergence though

Answer 7

may be lets work an example, are you working on anything specific ?

Answer 8

Yeah. Determine whether the series converges absolutely, conditionally, or diverges. I already proved that the ratio test diverged. \[\sum_{n=1}^{\infty}(-1)^{n+1}((1+n)/n^2)\]

Answer 9

And I started using the alternating series test, but I didn't know if that showed that the series converged conditionally, or converged in general

Answer 10

again, alternating series test has nothing to do with "conditional convergence"

To say the series is "conditionally convergent", you must establish two things :
1) the actual series converges.
2) the absolute value of series diverges

Answer 11

Oooh ok, right. I think that I understand what my teacher meant now when she said that it proved conditional convergence, because we made a flowchart. You could only use the alternating series test if absolute convergence didn't work out

Answer 12

But in itself, it doesn't actually prove conditional convergence, just that the series converged

Answer 13

So because the series that I posted diverged when tested for absolute convergence, and converged when tested with the AST, it converges conditionally

Answer 14

Yes..

\[\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1+n}{n^2}\]

\(b_n := \frac{1+n}{n^2}\)
1) Clearly \(b_n\ge 0\) for all \(n\).

2) the sequence \(\{b_n\}\) is decreasing because :
\[\color{blue}{b_{n+1}}=\frac{n+1+1}{(n+1)^2} = \frac{1}{n+1}+\frac{1}{(n+1)^2} \color{blue}{\lt} \frac{1}{n}+\frac{1}{n^2}=\frac{1+n}{n^2}=\color{blue}{b_n}\]

3) \(\lim\limits_{n\to\infty}b_n = \lim\limits_{n\to\infty}\frac{1+n}{n^2}=0\)

Therefore this series meets the hypotheiss of alternating series test, so the series \(\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1+n}{n^2}\) converges by alternating series test.

Answer 15

Alternating series test story ends there.

Answer 16

Notice that you need to still test the convergence of absolute value of series to say whether it covnerges absolutely or conditionally

Answer 17

Yes, and that diverged, so it converges conditionally?

Answer 18

Yes!

Answer 19

Ok great! I also had a quick question about the ratio test if you dont mind

Answer 20

sure ask

Answer 21

Ok. Do I always need the absolute value bars for the ratio test?

Answer 22

Yes, thats trur for ratio test also

Answer 23

Ok, so even if the series could not possibly have any negative terms (for example, 1/n), you still need the abs vals?

Answer 24

absolute value of a positive term is positive, so it wont change in anything. but it is good to put absolute bars always when u apply ratio test

Answer 25

|1/n| = 1/n for n>0

Answer 26

So the ratio test always proves absolute convergence regardless of the sign of the terms in the series

Answer 27

that looks like a correct statement

Answer 28

Ok. Thank you so much!! I might have to get a tattoo that says "the alternating series test only proves regular convergence"

Answer 29

Wait, then why don't we start with the alternating series test if that proves convergence? Absolute convergence tests dont matter if the series doesn't converge at all

Answer 30

Alternating series test is the first thing to try if you see (-1)^n in your series

Answer 31

If it doesn't have (-1)^n, how do I prove that it just generally converges?

Answer 32

use any other tests

Answer 33

I just found a series that does not follow the AST, but converges absolutely as well

Answer 34

possible, what is it ?

Answer 35

(-1)^n+1 (sin n/n^2)

Answer 36

Doesn't that mean that it doesn't converge, but it converges absolutely? That doesn't make any sense

Answer 37

did you use comparison with p-series to determine it converges absolutely ?

Answer 38

Yes

Answer 39

the LCT

Answer 40

good, since the absolute value of series converges, the actual series also converges.

Answer 41

But it doesn't pass the AST, so that means that it doesn't converge

Answer 42

absolute convergence =====> the actual series converge

Answer 43

who cares about alternating series test

Answer 44

I thought that you said the AST proves convergence

Answer 45

you said we cannot apply alternating series test here because the terms are not positive

Answer 46

so we're done with alternating series test. it is silent for this series.

Answer 47

The terms aren't all positive. Theres a (-1)^n+1

Answer 48

and a sin(n)

Answer 49

yes bn = sin(n)/n^2

the terms in bn are not all positive, so we cannot apply alternating series test

Answer 50

that doesn't mean the series diverges.
that just means alternating series test is silent and we need to try some other test

Answer 51

But I think that I understand. You can only use the AST if absolute convergence doesn't work. The fact that absolute convergence didn't work means that if the AST works, it has to be conditional convergence! I think that I get it now!

Answer 52

Does that sound legit?

Answer 53

that looks okay for now, but after working a few more problems you will see why i said "absolute convergence/conditional convergence" has nothing to do with AST.

Answer 54

I think that I understand that too. AST proves that the series has to converge, and the only other way that it can converge is conditional