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Hitaro9
 one year ago
Let ak >=0. Let the series ak diverge. Prove that the series ak/(1+ak) diverges
Hitaro9
 one year ago
Let ak >=0. Let the series ak diverge. Prove that the series ak/(1+ak) diverges

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hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1There's nothing super simple like a comparison that can be made as far as I can see. How would one go about proving this sort of question? Do you get in to definitions of series divergence or?

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{k=1}^{\infty}a _{k}\div(a _{k}+1)\]

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Hey, last problem. Thanks for your help on the last one @IrishBoy123. I'm just not sure what the general strategy is here.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1compare like terms in each series

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think you may simply use the limit test : \(\lim\limits_{k\to\infty} \dfrac{a_k}{1+a_k} \ne 0 \implies \sum\limits_{k=0}^{\infty}\dfrac{a_k}{1+a_k} \text{ diverges}\)

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Huh. How do we know ak / 1+ak doesn't go to zero? Couldn't ak go to zero and not converge (ie, 1/k)? Or do I have something off here?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yeah, tad confused here too we do not know that \(\lim\limits_{k\to\infty} a_k \ne 0\) as \(a_k\) might have failed that test and still diverged, ie it is poss that \(\lim\limits_{k\to\infty} a_k = 0\) plus \(\dfrac{\dfrac{a_k}{1+a_k}}{a_k} = \dfrac{1}{1 + a_k}\) and \(a_k >=0\) so the comparison test draws a blank

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$ \sum \frac{\frac 1 n }{1 + \frac 1 n }= \sum \frac {1}{n+1 }= \infty $$

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3If \(\lim\limits_{k\to\infty} a_k = 0\), then limit comparison test does the job : \(\lim\limits_{k\to\infty} \dfrac{\dfrac{a_k}{1+a_k}}{a_k} = 1\) therefore both the series, \(\sum a_k \) and \(\sum\dfrac{a_k}{1+a_k}\) will have same fate.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks im just repeating what irishboy has said

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1no, i chickened out of your second conclusion. can't remember why now :)

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1So why does 1/1+ak > that it diverges exactly? Sorry if this is obvious.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3hmm question is not so clear... could you elaborate a bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0limit comparison test: http://prntscr.com/8j65pv

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Oh wait right. Thank you Jayz. Okay then that makes sense.

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1So basically we have two cases, one for which ak > 0 and one for otherwise? If I understand correctly?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:) no problem. ganeshie and irishboy did the hard work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh you mean if a_k converged, then yes

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Okay yeah, my question was poorly phrased. I think I understand now. I'll go through and make it more wordy with respects to the limit test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We get two statements for free using limit comparison. 1. Let ak >=0. Let the series ak diverge. Then the series ak/(1+ak) diverges 2. 1. Let ak >=0. Let the series ak converge. Then the series ak/(1+ak) converges

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Thanks all of you for the help. I appreciate it.

hitaro9
 one year ago
Best ResponseYou've already chosen the best response.1Alright, I gotta get to class, thanks again for all your help. You're all amazing people.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just to go over the argument one more time to make this proof airtight. We are given ak>=0 and ∑ak diverges. Show that ∑ak / (1 + ak) diverges. proof: there are two cases lim ak ≠0 or lim ak = 0 1. If lim ak ≠0 , then since ak>=0 we have lim ak/(1 + ak) ≠0 and thus ∑ak / (1 + ak) diverges by divergence test. 2. If lim ak = 0 then lim [ ak / (1+ ak) ] / ak = lim ( 1 / (1 + ak) = 1 / (1 + 0) = 1 . by limit comparison test ∑ak / (1 + ak) diverges since ∑ak diverges
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