Question

Does the series (below) converge absolutely, conditionally, or diverge?

Answer 1

\[\sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n}\]

Answer 2

I know that it converges absolutely, but doesn't it fill all 3 criteria for the alternating series test?

Answer 3

I have checked online and people are only mentioning the absolute convergence

Answer 4

The answer seems to only be that it converges absolutely, but it also fulfills all 3 criteria for the alternating series test, so shouldn't it converge conditionally as well?

Answer 5

if the absolute value of series converges, then the actual series also converges.

Answer 6

Yes, I know, but can't a series converge conditionally as well?

Answer 7

If a series converges absolutely, then a series converges conditionally.
the converse is not always true (e.g. \( \sum \)(-1)^n * 1/n )
therefore absolute convergence is a stronger condition

Answer 8

we say a series converges conditionally when the absolute value of series diverges but the actual series converges

Answer 9

So I dont even need to prove that it converges conditionally if it converges absolutely?

Answer 10

$$ \Large \rm \sum |a_n|~ converges \Rightarrow \sum a_n ~converges


$$

Answer 11

I'm talking about conditionally. If a series converges absolutely then it converges conditionally as well?

Answer 12

yes

Answer 13

But some of the problems only converge absolutely

Answer 14

\[\sum\limits_{n=1}^{\infty} |a_n| \ge \sum\limits_{n=1}^{\infty} a_n\]

If the left side series converges, then by comparison test the right side series also converges.

Answer 15

I know that. My confusion is with conditional convergence

Answer 16

@perl some of the series converge absolutely but not conditionally, so thats not true

Answer 17

perl is right, above comparison test tells you that if a series converges absolutely, then the actual series also converges.

Answer 18

conditionally

Answer 19

I am talking about conditional convergence

Answer 20

you seem to have two confusions regarding terminology, let me clear them up

Answer 21

conditional convergence is not regular convergence

Answer 22

And Perl said that if a function converges absolutely, then it converges conditionally as well, which I disproved in my homework like 3 times

Answer 23

could you show one example of your homework where you proved absolute convergence need not imply conditional convergence ?

Answer 24

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

Answer 25

sin(n)/n^2 is not always positive, so it does not pass the alternating series test

Answer 26

But it passes the limit comparison test

Answer 27

You're correct about that series not passing the hypothesis of alternating series test.

So that just means you cannot use alternating series test for that series. Thats all.

Answer 28

We haven't been given any other tests for conditional convergence

Answer 29

Alternating series test never tells you about divergence

Answer 30

But it does tell you whether a series converges conditionally or not

Answer 31

You cannot apply alternating series test here because the series doesn't meet the hypothesis of alternating series test. End of story.

Answer 32

go and use some other test

Answer 33

I wasn't given another conditional convergence test

Answer 34

That was THE only one

Answer 35

lets look at absolute value of series first

Answer 36

Ok

Answer 37

\[\sum_{n=1}^{\infty}\left| (-1)^{n+1}\frac{ \sin(n) }{ n^2 }\right|~~ =~~ \sum_{n=1}^{\infty}\left|\frac{ \sin(n) }{ n^2 }\right| \]

Answer 38

Yeah, and that converges absolutely by the limit comparison test

Answer 39

Since \(|\sin(n)|\le 1\), we have \[\sum_{n=1}^{\infty}\left|\frac{ \sin(n) }{ n^2 }\right| \le \sum_{n=1}^{\infty}\frac{ 1 }{ n^2 }\]

By comparison with p-series test the left hand side series converges.

Answer 40

The absolute value of series converges, therefore the series also converges. And we say the series converges absolutely.

Answer 41

Ok

Answer 42

Absolute converges ====> actual series converges

Answer 43

So therefore it is supposed to converge conditionally, right?

Answer 44

Absolute convergence : when the "absolute value of series" converges, the "actual series" also converges and we say the series converges absolutely.

Conditional convergence : when the "absolute value of series" diverges, but the "actual series" converges. we say the series only converges conditionally

Answer 45

mmm

Answer 46

it is wrong to say
Absolute convergence ====> conditional convergence


correct way to say is :
Absolute convergence ====> the actual series converges

Answer 47

Perl said that the first one is true

Answer 48

Stick to the definitions and they will make sense

Answer 49

Ok. But a series can converge both absolutely AND conditionally, right?

Answer 50

tell me this, when do you say "a series converges conditionally" ?

Answer 51

also provide one example of a series that converges conditionally

Answer 52

A series converges conditionally if it passes the Alternating Series Test (it can fill the 3 criteria for the test)

Answer 53

\[\sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^n\] converges conditionally

Answer 54

Wrong.

Answer 55

How so

Answer 56

To say "a series converges conditionally", you must do two things :

1) you must prove that the "absolute value of series diverges"
2) the actual series converges.

Answer 57

does the abolute value of your example series diverge ?

Answer 58

Taken directly from my book: "The series \[\sum_{n=1}^{\infty}(-1)^n+1Un \] converges if all three of the following conditions are satisfied:
1. Each Un is positive;
2. Un >= Un+1 for all n>=N, for some integer N;
3. limn->infty Un->0

Answer 59

It should be (-1)^(n+1)Un

Answer 60

yeah thats alternating series test

Answer 61

Which proves conditional convergence

Answer 62

which can be used to test convergence of "actual series"

Answer 63

Oh snap, I just found a small side-note section talking about conditional convergence only happening if it isn't absolutely convergent, but still convergent

Answer 64

To say "the seris converges conditionally", you must show that the "absolute value of series diverges"

Answer 65

Yep thats right

Answer 66

Oh my god. That means that I just spent like 2 hours doing my homework wrong

Answer 67

In my defense, my teacher never made that clear to us

Answer 68

Haha happens, this thing trips everyone in the start

Answer 69

Ok. So I have to use like the ratio test or the LCT or the DCT or etc to show that the absolute value doesn't converge??

Answer 70

which series are you talking about

Answer 71

Well any series if I want to prove that it converges conditionally (or doesn't if the absolute value doesn't diverge)

Answer 72

Yes. lets work a quick example

Answer 73

Ok cool

Answer 74

\[\sum\limits_{n=1}^{\infty}\dfrac{(-1)^n}{n}\]

state whether the series is :
1) conditionally convergent, or
2) absolutely convergent

Answer 75

It can technically be both though, right?

Answer 76

Not specifically this one, but generally?

Answer 77

it can never be both.

Answer 78

look at definition of each of those terms again

Answer 79

Not on the same interval but I mean if a series has an interval of convergence, it could be conditionally convergent at the endpoints but absolutely convergent in the middle

Answer 80

lets not even talk about power series/interval of convergence yet

Answer 81

Ok, but I think that's the root of my confusion

Answer 82

the definitions of "absolute convergence" or "conditional convergence" have nothing to do with power series / alternating series

Answer 83

the definitions are more general, they are not based on alternating series test

Answer 84

you can tell whether a series is "absolutely convergenct" or "conditionally convergent" without using alternating series test.

Answer 85

Ok. But it is possible that if a series has an interval of convergence, that it exhibits both types of convergence. Not in the same places, but over the interval, both happen at different times. I got that the absolute value diverges BTW

Answer 86

Yes, in any given interval, a series is EITHER "absolutely convergent" or "conditionally convergent".

It is wrong to say the series is both "absolutely convergent and conditionally convergent". It makes no sense