anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
hell
anonymous
  • anonymous
if series |ak| converges then lim |ak|=0, then lim ak=0 , fine. then what does that show

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anonymous
  • anonymous
i didnt mean to write hell
anonymous
  • anonymous
that's alright
anonymous
  • anonymous
If lim ak = 0 that gets us started
anonymous
  • anonymous
i meant to write hello
anonymous
  • anonymous
because now it rules out ak diverging for sure and we are left with having to show that it converges
anonymous
  • anonymous
ok
anonymous
  • anonymous
so we have shown that lim ak = 0 , and by contradiction assumption we have that series ak diverges
anonymous
  • anonymous
we may or may not need contradiction, I haven't actually done out the whole proof, I'm just thinking out loud
anonymous
  • anonymous
because you might be able to use one of the tests for convergence (maybe the alternating series test) to show convergence
anonymous
  • anonymous
oh , we have two possibilities, either series |ak| = series ak, or it doesnt (by a negative factor)
anonymous
  • 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
  • anonymous
I may have another approach to the proof to share.
anonymous
  • 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
  • anonymous
no i was wrong about that, you can have series |ak| not equal to series ak but series ak is not alternating.
anonymous
  • 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
  • anonymous
ok where ak is a term in the sequence
anonymous
  • anonymous
ok so we have series [ ak + |ak| ] < = series 2 |ak|
anonymous
  • anonymous
because 0 <= [ ak + |ak| ] < = 2 |ak| can you sum both sides ?
anonymous
  • anonymous
yes we can so we have series ak + series |ak| < = series 2|ak|
anonymous
  • anonymous
subtract series |ak| from both sides
anonymous
  • anonymous
so series ak < series |ak| , and if series |ak| diverges than series ak must diverge , contradiction
anonymous
  • 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
  • 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
  • anonymous
You want to prove convergence?
anonymous
  • anonymous
yes but we already assumed ak is convergent, we want to show a contradiction
anonymous
  • anonymous
oh youre doing a direct proof
anonymous
  • anonymous
you actually dont need a contradiction here
anonymous
  • anonymous
I only thought it would help just by taking a quick look at the problem but the way AnwarA posted works better
anonymous
  • anonymous
I really don't understand, What is it exactly that you want to prove?
anonymous
  • anonymous
oh
anonymous
  • anonymous
sorry we were approaching this differently, you proved it
anonymous
  • anonymous
isnt there a simpler proof, here
anonymous
  • 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
  • anonymous
how did you get this
anonymous
  • anonymous
Get what?
anonymous
  • anonymous
series ak = series [ (ak + |ak| ) - series |ak |
anonymous
  • anonymous
so my proof doesnt work ?
anonymous
  • 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
  • anonymous
can you tiny url that
anonymous
  • anonymous
if you copy and paste it you should get it
anonymous
  • anonymous
It discusses absolute convergence and the first part is the proof AnwarA just posted.
anonymous
  • anonymous
so you did series ak = series [ ak + |ak| - |ak| ] , and then you split that series on the right
anonymous
  • anonymous
but theres some condition about splitting terms in series
anonymous
  • anonymous
series (an + bn ) = series an + series bn , as long as ...
anonymous
  • anonymous
as long as series an and series bn converge
anonymous
  • anonymous
sorry you might need to refresh openstudy.com
anonymous
  • 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
  • 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
  • anonymous
suppose you have series [ 2/n - 1/n] = series 2/n - series 1/n ?
anonymous
  • anonymous
infinity - infinity is not determinate
anonymous
  • anonymous
im trying to think of a counterexample, where series (an + bn) != series an + series bn
anonymous
  • anonymous
like an infinite case
anonymous
  • anonymous
You're confusing me a little bit :).. These two things are totally different, I mean your last three replies.
anonymous
  • anonymous
well linearity of series works only under certain conditions
anonymous
  • anonymous
So you're asking if series [ak+an]=series [ak]+series [an]?
anonymous
  • anonymous
right, you implicitly used that , under what conditions is that true
anonymous
  • anonymous
Hmm. I don't think there are any conditions.
anonymous
  • anonymous
ok , how do you prove that lim |an| = 0 iff lim an = 0
anonymous
  • anonymous
Oh wait, It's true for ak and an convergent series.
anonymous
  • anonymous
ahhh
anonymous
  • anonymous
since you might get funny results with divergent series
anonymous
  • 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
  • anonymous
if we can show series |ak| converges, we get series ak converging for free
anonymous
  • 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
  • anonymous
Yeah, you're right.
anonymous
  • anonymous
about which part
anonymous
  • anonymous
oh the stronger aspect
anonymous
  • anonymous
Yeah.. Let me finish the first part of the proof :)
anonymous
  • anonymous
sure, sorry for interrupting. youre the best man
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
if an <0 then ?
anonymous
  • anonymous
youre proving that if lim |an| = 0 -> lim an = 0
anonymous
  • 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
  • anonymous
but you can have a mixture of positive and negative
anonymous
  • anonymous
like alternating , such as a1 >0 , a2 < 0 , etc
anonymous
  • 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
  • anonymous
oh
anonymous
  • anonymous
Do you see it? Try reading it again.
anonymous
  • anonymous
yes but an can alternate
anonymous
  • anonymous
An
anonymous
  • anonymous
Yeah, that makes no difference since you ALREADY KNOW that lim |an| is zero.
anonymous
  • anonymous
but that*
anonymous
  • anonymous
i though you said we assume lim an = 0
anonymous
  • anonymous
and we want to prove lim |an | = 0
anonymous
  • anonymous
We are not assuming that, the question is. GO and read your question again.
anonymous
  • anonymous
ok theres too parts
anonymous
  • anonymous
assuming lim |an| = 0, then show lim an = 0 the other one is assuming lim an = 0 show lim |an | = 0
anonymous
  • anonymous
Exactly :)
anonymous
  • anonymous
so which one did you prove
anonymous
  • anonymous
We have done the second part, you should try the first one.
anonymous
  • anonymous
but youre mixing up specific ak and a general ak
anonymous
  • anonymous
What do you mean?
anonymous
  • anonymous
well for example , take (-1)^n / n^2 , the sequence
anonymous
  • anonymous
Ok?
anonymous
  • anonymous
so you said that either |an| = an or | an | = -an, but in this case its neither
anonymous
  • anonymous
Yeah, this is the definition of absolute value.
anonymous
  • anonymous
no its neither
anonymous
  • anonymous
Are you sure?
anonymous
  • anonymous
here | (-1)^n / n^2 ) | != (-1)^n / n^2, and | (-1)^n / n^2 | != - (-1)^n / n^2
anonymous
  • anonymous
where != means not equal
anonymous
  • anonymous
so your statement or proof is incorrect
anonymous
  • anonymous
| (-1)^n / n^2 ) |= 1 / n^2
anonymous
  • anonymous
an = (-1)^n / n^2, |an| is not equal to -an , nor an .
anonymous
  • anonymous
:) I said for an>=0 |an|=an, and for an<0 |an|=-an. This is what absolute value is.
anonymous
  • anonymous
yes and ?
anonymous
  • anonymous
we have to show for all lim |an| = 0 , how are you going to put those two things together
anonymous
  • anonymous
wait so youre splitting up the limit ?
anonymous
  • anonymous
It's clear!! the limit of the negative part is zero, and so is the positive part. right?
anonymous
  • anonymous
what do you mean
anonymous
  • anonymous
oh yes
anonymous
  • anonymous
the limit of the negative terms is 0
anonymous
  • anonymous
Got it?
anonymous
  • anonymous
but how do you show the whole limit is going to zeor
anonymous
  • anonymous
zero
anonymous
  • anonymous
you showed that some of the terms go to zero, and some of the terms term go to zero
anonymous
  • anonymous
Limit is a linear operation, isn't it?
anonymous
  • anonymous
so the positive terms go to zero, true, and the negative terms go to zero, true , and?
anonymous
  • anonymous
so you split An into positive and negative terms ?
anonymous
  • anonymous
You can say so.
anonymous
  • anonymous
but thats not rigorous
anonymous
  • anonymous
maybe a contradiction proof
anonymous
  • anonymous
but graphically it makes sense
anonymous
  • anonymous
if we graph An = f(n) on the x y plane
anonymous
  • anonymous
the negative terms are below the x axis
anonymous
  • anonymous
and they are going to zero
anonymous
  • anonymous
the positive terms above the x axis are going to zero
anonymous
  • anonymous
not rigorous? show me what a rigorous proof is.
anonymous
  • anonymous
err, not deductive
anonymous
  • anonymous
ok i disproved your proof
anonymous
  • anonymous
simple counterexample
anonymous
  • anonymous
are you ready
anonymous
  • anonymous
Lol, It's a "proof", you can't disprove it. But sure, show me what you got :)
anonymous
  • anonymous
what if an = (-1)^n / n^2
anonymous
  • anonymous
no it isnt
anonymous
  • anonymous
its not a proof
anonymous
  • anonymous
you wrote that lim |an| = lim an or lim |an| = - lim an , FALSE
anonymous
  • anonymous
lim |an | = lim an or that lim |an | = lim -an , and i showed you a counterexample
anonymous
  • anonymous
Yeah?! :)
anonymous
  • anonymous
thats what you wrote, its a false statement
anonymous
  • anonymous
Claim: lim |an | = lim an or that lim |an | = lim (-an) False,. let an = (-1)^n / n^2
anonymous
  • anonymous
lim | (-1)^n / n^2 | is not lim (-1)^n / n^2 and it is not lim - (-1)^n / n^2
anonymous
  • anonymous
It's lim an when an>=0 and it's lim -an when an<0.
anonymous
  • anonymous
an = (-1)^n / n^2 lim |an| is not equal to lim an, and lim |an| is not equal to lim (-an)
anonymous
  • anonymous
and we need to show lim |an| = 0 for all an
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
you didnt show for all an, you showed that if an is always positive, or if an is always negative
anonymous
  • anonymous
You didn't even bother to understand what I wrote.
anonymous
  • anonymous
sure i did, relax for a second,
anonymous
  • anonymous
I am all relaxed.
anonymous
  • 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
  • anonymous
THATS what you proved
anonymous
  • anonymous
Is it?
anonymous
  • anonymous
you proved two seperate statements, and thats NOT the conclusion
anonymous
  • anonymous
yup, go back and think about it
anonymous
  • 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
  • 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
  • anonymous
and my counterexample blows your proof to pieces
anonymous
  • anonymous
an = (-1)^n/n^2, your proof does not handle such a creature
anonymous
  • anonymous
Haha
anonymous
  • anonymous
Is this a college calculus course?
anonymous
  • anonymous
Here is another proof then.
anonymous
  • anonymous
no, just me thinking
anonymous
  • anonymous
good thing you didnt freak out and starting calling me names , thanks
anonymous
  • anonymous
some people cant handle the pressure,
anonymous
  • anonymous
i like to know all the theorems relevant to a course, like calculus. and i like to see their proofs, if possible
anonymous
  • anonymous
If lim |an|=0, then -lim|an|=0, and since -|an|<=an<=|an|, lim an=0 by the squeeze theorem.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Is that a "proof"?
anonymous
  • anonymous
the other direction is tougher
anonymous
  • anonymous
YES , lol
anonymous
  • anonymous
you done good
anonymous
  • anonymous
:) I won't call you names other than cartorest (I don't know if that's even a name:P)
anonymous
  • anonymous
no i meant, you didnt freak out when i challenged your proof
anonymous
  • anonymous
Haha, why should I freak out? I like that actually.
anonymous
  • anonymous
:)
anonymous
  • anonymous
Good luck in your study. I had fun "Studying" with you tonight.
anonymous
  • anonymous
so , prove if lim an = 0 then lim |an | = 0
anonymous
  • anonymous
wait
anonymous
  • anonymous
the entree, the main course,
anonymous
  • anonymous
Yeah?
anonymous
  • anonymous
whats the proof , you proved the converse
anonymous
  • anonymous
I don't have another proof other than what I did before right now.
anonymous
  • anonymous
you mean the non proof
anonymous
  • anonymous
ok i have an idea. lets use contradiction, assume lim an = 0 and lim |an| != 0
anonymous
  • anonymous
i dont know
anonymous
  • 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
  • anonymous
i mean, my bad
anonymous
  • anonymous
I don't know either. I am so sleepy!!
anonymous
  • anonymous
you proved lim |an| = 0 then lim an = 0 .
anonymous
  • anonymous
with the sandwich theorem, but the converse we didnt prove yet
anonymous
  • anonymous
I have to go now, I'll catch up with you tomorrow.
anonymous
  • anonymous
Bye!! :)
anonymous
  • anonymous
ok , i just wanted to be clear we are in agreement
anonymous
  • anonymous
You really are funny.
anonymous
  • anonymous
well unlike religious views, math people MUST agree on things
anonymous
  • anonymous
i hope thats a compliment
anonymous
  • anonymous
It is :)
anonymous
  • anonymous
alright, maybe we will chat tomorrow , adios

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