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

- anonymous

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- anonymous

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)

- anonymous

hell

- anonymous

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

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## More answers

- anonymous

i didnt mean to write hell

- anonymous

that's alright

- anonymous

If lim ak = 0 that gets us started

- anonymous

i meant to write hello

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I may have another approach to the proof to share.

- anonymous

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 ...

- anonymous

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

- anonymous

|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.

- anonymous

ok where ak is a term in the sequence

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

subtract series |ak| from both sides

- anonymous

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

- anonymous

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.

- anonymous

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 ?

- anonymous

You want to prove convergence?

- anonymous

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

- anonymous

oh youre doing a direct proof

- anonymous

you actually dont need a contradiction here

- anonymous

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

- anonymous

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

- anonymous

oh

- anonymous

sorry we were approaching this differently, you proved it

- anonymous

isnt there a simpler proof, here

- anonymous

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 ,

- anonymous

how did you get this

- anonymous

Get what?

- anonymous

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

- anonymous

so my proof doesnt work ?

- anonymous

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

- anonymous

can you tiny url that

- anonymous

if you copy and paste it you should get it

- anonymous

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

- anonymous

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

- anonymous

but theres some condition about splitting terms in series

- anonymous

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

- anonymous

as long as series an and series bn converge

- anonymous

sorry you might need to refresh openstudy.com

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

infinity - infinity is not determinate

- anonymous

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

- anonymous

like an infinite case

- anonymous

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

- anonymous

well linearity of series works only under certain conditions

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ahhh

- anonymous

since you might get funny results with divergent series

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

Yeah, you're right.

- anonymous

about which part

- anonymous

oh the stronger aspect

- anonymous

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

- anonymous

sure, sorry for interrupting. youre the best man

- anonymous

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

- anonymous

ok

- anonymous

if an <0 then ?

- anonymous

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

- anonymous

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

- anonymous

but you can have a mixture of positive and negative

- anonymous

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

- anonymous

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.

- anonymous

oh

- anonymous

Do you see it? Try reading it again.

- anonymous

yes but an can alternate

- anonymous

An

- anonymous

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

- anonymous

but that*

- anonymous

i though you said we assume lim an = 0

- anonymous

and we want to prove lim |an | = 0

- anonymous

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

- anonymous

ok theres too parts

- anonymous

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

- anonymous

Exactly :)

- anonymous

so which one did you prove

- anonymous

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

- anonymous

but youre mixing up specific ak and a general ak

- anonymous

What do you mean?

- anonymous

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

- anonymous

Ok?

- anonymous

so you said that either |an| = an or | an | = -an, but in this case its neither

- anonymous

Yeah, this is the definition of absolute value.

- anonymous

no its neither

- anonymous

Are you sure?

- anonymous

here | (-1)^n / n^2 ) | != (-1)^n / n^2,
and | (-1)^n / n^2 | != - (-1)^n / n^2

- anonymous

where != means not equal

- anonymous

so your statement or proof is incorrect

- anonymous

| (-1)^n / n^2 ) |= 1 / n^2

- anonymous

an = (-1)^n / n^2,
|an| is not equal to -an , nor an .

- anonymous

:) I said for an>=0 |an|=an, and for an<0 |an|=-an. This is what absolute value is.

- anonymous

yes and ?

- anonymous

we have to show for all
lim |an| = 0 , how are you going to put those two things together

- anonymous

wait so youre splitting up the limit ?

- anonymous

It's clear!! the limit of the negative part is zero, and so is the positive part. right?

- anonymous

what do you mean

- anonymous

oh yes

- anonymous

the limit of the negative terms is 0

- anonymous

Got it?

- anonymous

but how do you show the whole limit is going to zeor

- anonymous

zero

- anonymous

you showed that some of the terms go to zero, and some of the terms term go to zero

- anonymous

Limit is a linear operation, isn't it?

- anonymous

so the positive terms go to zero, true, and the negative terms go to zero, true , and?

- anonymous

so you split An into positive and negative terms ?

- anonymous

You can say so.

- anonymous

but thats not rigorous

- anonymous

maybe a contradiction proof

- anonymous

but graphically it makes sense

- anonymous

if we graph An = f(n) on the x y plane

- anonymous

the negative terms are below the x axis

- anonymous

and they are going to zero

- anonymous

the positive terms above the x axis are going to zero

- anonymous

not rigorous? show me what a rigorous proof is.

- anonymous

err, not deductive

- anonymous

ok i disproved your proof

- anonymous

simple counterexample

- anonymous

are you ready

- anonymous

Lol, It's a "proof", you can't disprove it. But sure, show me what you got :)

- anonymous

what if an = (-1)^n / n^2

- anonymous

no it isnt

- anonymous

its not a proof

- anonymous

you wrote that lim |an| = lim an or lim |an| = - lim an , FALSE

- anonymous

lim |an | = lim an or that lim |an | = lim -an , and i showed you a counterexample

- anonymous

Yeah?! :)

- anonymous

thats what you wrote, its a false statement

- anonymous

Claim:
lim |an | = lim an or that lim |an | = lim (-an)
False,.
let an = (-1)^n / n^2

- anonymous

lim | (-1)^n / n^2 | is not lim (-1)^n / n^2
and it is not lim - (-1)^n / n^2

- anonymous

It's lim an when an>=0 and it's lim -an when an<0.

- anonymous

an = (-1)^n / n^2
lim |an| is not equal to lim an,
and lim |an| is not equal to lim (-an)

- anonymous

and we need to show lim |an| = 0 for all an

- anonymous

Yeah.

- anonymous

you didnt show for all an, you showed that if an is always positive, or if an is always negative

- anonymous

You didn't even bother to understand what I wrote.

- anonymous

sure i did, relax for a second,

- anonymous

I am all relaxed.

- anonymous

you proved that lim |an| goes to zero if an is always positive, and you proved lim |an| goes to zero when an is always negative

- anonymous

THATS what you proved

- anonymous

Is it?

- anonymous

you proved two seperate statements, and thats NOT the conclusion

- anonymous

yup, go back and think about it

- anonymous

do you remember your proof? you proved that if an is positive, so and so, if an is negative so and so. thats all. what if an is positive AND negative

- anonymous

if an is positive lim |an| = 0 , fine.
if an is negative lim |an| = 0. great, and? what if an is both positive and negative

- anonymous

and my counterexample blows your proof to pieces

- anonymous

an = (-1)^n/n^2, your proof does not handle such a creature

- anonymous

Haha

- anonymous

Is this a college calculus course?

- anonymous

Here is another proof then.

- anonymous

no, just me thinking

- anonymous

good thing you didnt freak out and starting calling me names , thanks

- anonymous

some people cant handle the pressure,

- anonymous

i like to know all the theorems relevant to a course, like calculus. and i like to see their proofs, if possible

- anonymous

If lim |an|=0, then -lim|an|=0, and since -|an|<=an<=|an|, lim an=0 by the squeeze theorem.

- anonymous

ok

- anonymous

Is that a "proof"?

- anonymous

the other direction is tougher

- anonymous

YES , lol

- anonymous

you done good

- anonymous

:) I won't call you names other than cartorest (I don't know if that's even a name:P)

- anonymous

no i meant, you didnt freak out when i challenged your proof

- anonymous

Haha, why should I freak out? I like that actually.

- anonymous

:)

- anonymous

Good luck in your study. I had fun "Studying" with you tonight.

- anonymous

so , prove if lim an = 0 then lim |an | = 0

- anonymous

wait

- anonymous

the entree, the main course,

- anonymous

Yeah?

- anonymous

whats the proof , you proved the converse

- anonymous

I don't have another proof other than what I did before right now.

- anonymous

you mean the non proof

- anonymous

ok i have an idea. lets use contradiction, assume lim an = 0 and lim |an| != 0

- anonymous

i dont know

- anonymous

well ill post it , thanks for your help. just realize you did not prove that lim |an | = 0 implies that lim an = 0

- anonymous

i mean, my bad

- anonymous

I don't know either. I am so sleepy!!

- anonymous

you proved lim |an| = 0 then lim an = 0 .

- anonymous

with the sandwich theorem, but the converse we didnt prove yet

- anonymous

I have to go now, I'll catch up with you tomorrow.

- anonymous

Bye!! :)

- anonymous

ok , i just wanted to be clear we are in agreement

- anonymous

You really are funny.

- anonymous

well unlike religious views, math people MUST agree on things

- anonymous

i hope thats a compliment

- anonymous

It is :)

- anonymous

alright, maybe we will chat tomorrow , adios

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