Question

show or prove that if series |ak| converges then series ak converges.
Details: the converse is clearly false. for example series (-1)^n/n converges by AST but series 1/n diverges

Answer 1

Assume |ak| converges then assume ak will diverge and show that it doesn't. (this might or might not work but that's what I would try)

Answer 2

hell

Answer 3

if series |ak| converges then lim |ak|=0, then lim ak=0 , fine. then what does that show

Answer 4

i didnt mean to write hell

Answer 5

that's alright

Answer 6

If lim ak = 0 that gets us started

Answer 7

i meant to write hello

Answer 8

because now it rules out ak diverging for sure and we are left with having to show that it converges

Answer 9

ok

Answer 10

so we have shown that lim ak = 0 , and by contradiction assumption we have that series ak diverges

Answer 11

we may or may not need contradiction, I haven't actually done out the whole proof, I'm just thinking out loud

Answer 12

because you might be able to use one of the tests for convergence (maybe the alternating series test) to show convergence

Answer 13

oh , we have two possibilities, either series |ak| = series ak, or it doesnt (by a negative factor)

Answer 14

if series |ak| = series ak then we have a contradiction.
if series |ak| != series ak , then the left side must be an alternating series?

Answer 15

I may have another approach to the proof to share.

Answer 16

either |ak| = ak, the sequence, or it doesnt. if they are not equal then ... oh i could be wrong about the alternating series. some series dont alternate like 1/3 + 1/5 - 1/7 - 1/9 ...

Answer 17

no i was wrong about that, you can have series |ak| not equal to series ak but series ak is not alternating.

Answer 18

|ak| is either -ak or ak, then we can write this inequality:
\[0\le a_k+\left| a_k \right|\le2 \left| a_k \right|\]
For instance, I want to show that ak + |ak| is convergent.

Answer 19

ok where ak is a term in the sequence

Answer 20

ok so we have series [ ak + |ak| ] < = series 2 |ak|

Answer 21

because 0 <= [ ak + |ak| ] < = 2 |ak|
can you sum both sides ?

Answer 22

yes we can so we have series ak + series |ak| < = series 2|ak|

Answer 23

subtract series |ak| from both sides

Answer 24

so series ak < series |ak| , and if series |ak| diverges than series ak must diverge , contradiction

Answer 25

First, 2|ak| is clearly a convergent series, since it's just a multiple of one. That implies that ak+|ak| is also a convergent by comparison. Having proved that, consider:
\[\sum_{}^{}a_k=\sum_{}^{}(a_k+\left| a_k \right|)-\sum_{}^{}\left| a_k \right|\]
Therefore, ak is the difference between two convergent series, and hence it's convergent.

Answer 26

oh, i think there is a simpler proof.
ak <= |ak| , and series ak <= series |ak| , but series |ak| diverges by reductio hypothesis , so series ak must diverge ?

Answer 27

You want to prove convergence?

Answer 28

yes but we already assumed ak is convergent, we want to show a contradiction

Answer 29

oh youre doing a direct proof

Answer 30

you actually dont need a contradiction here

Answer 31

I only thought it would help just by taking a quick look at the problem but the way AnwarA posted works better

Answer 32

I really don't understand, What is it exactly that you want to prove?

Answer 33

oh

Answer 34

sorry we were approaching this differently, you proved it

Answer 35

isnt there a simpler proof, here

Answer 36

ak < = |ak| , sum both sides
series ak <= series |ak|
by assumption series |ak| converges... so series ak must converge, oh but ak does not have to be greater than zero, i see ,

Answer 37

how did you get this

Answer 38

Get what?

Answer 39

series ak = series [ (ak + |ak| ) - series |ak |

Answer 40

so my proof doesnt work ?

Answer 41

http://docs.google.com/viewer?a=v&q=cache:9YPiblhNCroJ:math.furman.edu/~dcs/book/c5pdf/sec56.pdf+absolute+value+series+convergence&hl=en&gl=us&pid=bl&srcid=ADGEEShHfgxoxB6SJOaWlZWVBvNyVCFKytUuuleiuBd_O4ED-AnQTq2i6uMBbwkKZTu7JcRmZGPTAFieIoBUa3EvmfzN5pu8tkSGkIkxmS8MHLq9ymIaceiT7q2ler0q1gB2zKapveW9&sig=AHIEtbQu3f_nN9Psh11e12NfBXDXKS7MLQ

Answer 42

can you tiny url that

Answer 43

if you copy and paste it you should get it

Answer 44

It discusses absolute convergence and the first part is the proof AnwarA just posted.

Answer 45

so you did series ak = series [ ak + |ak| - |ak| ] , and then you split that series on the right

Answer 46

but theres some condition about splitting terms in series

Answer 47

series (an + bn ) = series an + series bn , as long as ...

Answer 48

as long as series an and series bn converge

Answer 49

sorry you might need to refresh openstudy.com

Answer 50

Well, it's clear:
\[\sum_{}^{}a_k=\sum_{}^{}(a_k+\left| a_k \right|)-\sum_{}^{}\left| a_k \right|=\sum_{}^{}a_k-\sum_{}^{}\left| a_k \right|+\sum_{}^{}\left| a_k \right|=\sum_{}^{}a_k\]

Answer 51

No i dont you can do that

my other question is , which kind of goes with this,
prove lim |an| = 0 iff lim an = 0

Answer 52

suppose you have series [ 2/n - 1/n] = series 2/n - series 1/n ?

Answer 53

infinity - infinity is not determinate

Answer 54

im trying to think of a counterexample,
where series (an + bn) != series an + series bn

Answer 55

like an infinite case

Answer 56

You're confusing me a little bit :).. These two things are totally different, I mean your last three replies.

Answer 57

well linearity of series works only under certain conditions

Answer 58

So you're asking if series [ak+an]=series [ak]+series [an]?

Answer 59

right, you implicitly used that , under what conditions is that true

Answer 60

Hmm. I don't think there are any conditions.

Answer 61

ok , how do you prove that lim |an| = 0 iff lim an = 0

Answer 62

Oh wait, It's true for ak and an convergent series.

Answer 63

ahhh

Answer 64

since you might get funny results with divergent series

Answer 65

ok to recap, we showed that absolute convergence is a stronger condition. since if series |ak| converges then series ak converges. but the converse is not necessarily true, so its called conditional convergence , ie series ak converges but |ak| does not converge

Answer 66

if we can show series |ak| converges, we get series ak converging for free

Answer 67

To prove that lim |an| = 0 iff lim an = 0, we have to do it in two direction (since it's iff statement). The first part is to prove that lim |an|=0 if lim an=0 as n approaches infinity.

Answer 68

Yeah, you're right.

Answer 69

about which part

Answer 70

oh the stronger aspect

Answer 71

Yeah.. Let me finish the first part of the proof :)

Answer 72

sure, sorry for interrupting. youre the best man

Answer 73

|an| is either an or -an. If an>=0, that's |an|=an, then:
\[\lim_{n \rightarrow \infty}\left| a_n \right|=\lim_{n \rightarrow \infty}a_n=0\]

Answer 74

ok

Answer 75

if an <0 then ?

Answer 76

youre proving that if lim |an| = 0 -> lim an = 0

Answer 77

If an<0 then |an|=-an, that's:
\[\lim_{n \rightarrow \infty}\left| a_n \right|=-\lim_{n \rightarrow \infty}a_n=0\]

Answer 78

but you can have a mixture of positive and negative

Answer 79

like alternating , such as a1 >0 , a2 < 0 , etc

Answer 80

No, I am proving the opposite. That's if I KNOW that lim an=0, then lim |an|=0. I took the two cases when an>=0 or an<0.

Answer 81

oh

Answer 82

Do you see it? Try reading it again.

Answer 83

yes but an can alternate

Answer 84

An

Answer 85

Yeah, that makes no difference since you ALREADY KNOW that lim |an| is zero.

Answer 86

but that*

Answer 87

i though you said we assume lim an = 0

Answer 88

and we want to prove lim |an | = 0

Answer 89

We are not assuming that, the question is. GO and read your question again.

Answer 90

ok theres too parts

Answer 91

assuming lim |an| = 0, then show lim an = 0
the other one is
assuming lim an = 0 show lim |an | = 0

Answer 92

Exactly :)

Answer 93

so which one did you prove

Answer 94

We have done the second part, you should try the first one.

Answer 95

but youre mixing up specific ak and a general ak

Answer 96

What do you mean?

Answer 97

well for example , take (-1)^n / n^2 , the sequence

Answer 98

Ok?