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anonymous
 5 years ago
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
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assume 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
 5 years ago
Best ResponseYou've already chosen the best response.0if series ak converges then lim ak=0, then lim ak=0 , fine. then what does that show

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt mean to write hell

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If lim ak = 0 that gets us started

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant to write hello

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because now it rules out ak diverging for sure and we are left with having to show that it converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have shown that lim ak = 0 , and by contradiction assumption we have that series ak diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we may or may not need contradiction, I haven't actually done out the whole proof, I'm just thinking out loud

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because you might be able to use one of the tests for convergence (maybe the alternating series test) to show convergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh , we have two possibilities, either series ak = series ak, or it doesnt (by a negative factor)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if series ak = series ak then we have a contradiction. if series ak != series ak , then the left side must be an alternating series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I may have another approach to the proof to share.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0either 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
 5 years ago
Best ResponseYou've already chosen the best response.0no i was wrong about that, you can have series ak not equal to series ak but series ak is not alternating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ak 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok where ak is a term in the sequence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so we have series [ ak + ak ] < = series 2 ak

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because 0 <= [ ak + ak ] < = 2 ak can you sum both sides ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes we can so we have series ak + series ak < = series 2ak

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0subtract series ak from both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so series ak < series ak , and if series ak diverges than series ak must diverge , contradiction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, 2ak 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You want to prove convergence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but we already assumed ak is convergent, we want to show a contradiction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh youre doing a direct proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you actually dont need a contradiction here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I only thought it would help just by taking a quick look at the problem but the way AnwarA posted works better

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I really don't understand, What is it exactly that you want to prove?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry we were approaching this differently, you proved it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt there a simpler proof, here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ak < = 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
 5 years ago
Best ResponseYou've already chosen the best response.0series ak = series [ (ak + ak )  series ak 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my proof doesnt work ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you tiny url that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you copy and paste it you should get it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It discusses absolute convergence and the first part is the proof AnwarA just posted.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you did series ak = series [ ak + ak  ak ] , and then you split that series on the right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but theres some condition about splitting terms in series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0series (an + bn ) = series an + series bn , as long as ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as long as series an and series bn converge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry you might need to refresh openstudy.com

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0No 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
 5 years ago
Best ResponseYou've already chosen the best response.0suppose you have series [ 2/n  1/n] = series 2/n  series 1/n ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0infinity  infinity is not determinate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to think of a counterexample, where series (an + bn) != series an + series bn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like an infinite case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're confusing me a little bit :).. These two things are totally different, I mean your last three replies.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well linearity of series works only under certain conditions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you're asking if series [ak+an]=series [ak]+series [an]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, you implicitly used that , under what conditions is that true

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm. I don't think there are any conditions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , how do you prove that lim an = 0 iff lim an = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh wait, It's true for ak and an convergent series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since you might get funny results with divergent series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0if we can show series ak converges, we get series ak converging for free

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh the stronger aspect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. Let me finish the first part of the proof :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure, sorry for interrupting. youre the best man

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an 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
 5 years ago
Best ResponseYou've already chosen the best response.0youre proving that if lim an = 0 > lim an = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If an<0 then an=an, that's: \[\lim_{n \rightarrow \infty}\left a_n \right=\lim_{n \rightarrow \infty}a_n=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you can have a mixture of positive and negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like alternating , such as a1 >0 , a2 < 0 , etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you see it? Try reading it again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but an can alternate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that makes no difference since you ALREADY KNOW that lim an is zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i though you said we assume lim an = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we want to prove lim an  = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We are not assuming that, the question is. GO and read your question again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0assuming lim an = 0, then show lim an = 0 the other one is assuming lim an = 0 show lim an  = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so which one did you prove

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We have done the second part, you should try the first one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but youre mixing up specific ak and a general ak

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well for example , take (1)^n / n^2 , the sequence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you said that either an = an or  an  = an, but in this case its neither

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, this is the definition of absolute value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here  (1)^n / n^2 )  != (1)^n / n^2, and  (1)^n / n^2  !=  (1)^n / n^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where != means not equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your statement or proof is incorrect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0 (1)^n / n^2 ) = 1 / n^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an = (1)^n / n^2, an is not equal to an , nor an .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) I said for an>=0 an=an, and for an<0 an=an. This is what absolute value is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to show for all lim an = 0 , how are you going to put those two things together

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait so youre splitting up the limit ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's clear!! the limit of the negative part is zero, and so is the positive part. right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limit of the negative terms is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how do you show the whole limit is going to zeor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you showed that some of the terms go to zero, and some of the terms term go to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Limit is a linear operation, isn't it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the positive terms go to zero, true, and the negative terms go to zero, true , and?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you split An into positive and negative terms ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but thats not rigorous

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe a contradiction proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but graphically it makes sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we graph An = f(n) on the x y plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the negative terms are below the x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and they are going to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the positive terms above the x axis are going to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not rigorous? show me what a rigorous proof is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i disproved your proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0simple counterexample

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, It's a "proof", you can't disprove it. But sure, show me what you got :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what if an = (1)^n / n^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you wrote that lim an = lim an or lim an =  lim an , FALSE

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim an  = lim an or that lim an  = lim an , and i showed you a counterexample

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what you wrote, its a false statement

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Claim: lim an  = lim an or that lim an  = lim (an) False,. let an = (1)^n / n^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim  (1)^n / n^2  is not lim (1)^n / n^2 and it is not lim  (1)^n / n^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's lim an when an>=0 and it's lim an when an<0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an = (1)^n / n^2 lim an is not equal to lim an, and lim an is not equal to lim (an)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we need to show lim an = 0 for all an

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you didnt show for all an, you showed that if an is always positive, or if an is always negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You didn't even bother to understand what I wrote.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure i did, relax for a second,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0THATS what you proved

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you proved two seperate statements, and thats NOT the conclusion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup, go back and think about it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do 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
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0and my counterexample blows your proof to pieces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an = (1)^n/n^2, your proof does not handle such a creature

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this a college calculus course?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here is another proof then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good thing you didnt freak out and starting calling me names , thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0some people cant handle the pressure,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i like to know all the theorems relevant to a course, like calculus. and i like to see their proofs, if possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If lim an=0, then liman=0, and since an<=an<=an, lim an=0 by the squeeze theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the other direction is tougher

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) I won't call you names other than cartorest (I don't know if that's even a name:P)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i meant, you didnt freak out when i challenged your proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha, why should I freak out? I like that actually.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck in your study. I had fun "Studying" with you tonight.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so , prove if lim an = 0 then lim an  = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the entree, the main course,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the proof , you proved the converse

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't have another proof other than what I did before right now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean the non proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i have an idea. lets use contradiction, assume lim an = 0 and lim an != 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well ill post it , thanks for your help. just realize you did not prove that lim an  = 0 implies that lim an = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know either. I am so sleepy!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you proved lim an = 0 then lim an = 0 .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with the sandwich theorem, but the converse we didnt prove yet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to go now, I'll catch up with you tomorrow.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , i just wanted to be clear we are in agreement

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You really are funny.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well unlike religious views, math people MUST agree on things

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i hope thats a compliment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, maybe we will chat tomorrow , adios
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