Hitaro9
  • Hitaro9
Let ak >=0. Let the series ak diverge. Prove that the series ak/(1+ak) diverges
Mathematics
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SOLVED
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katieb
  • katieb
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Hitaro9
  • Hitaro9
There'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
  • Hitaro9
\[\sum_{k=1}^{\infty}a _{k}\div(a _{k}+1)\]
Hitaro9
  • Hitaro9
|dw:1442920394346:dw|

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Hitaro9
  • Hitaro9
Hey, last problem. Thanks for your help on the last one @IrishBoy123. I'm just not sure what the general strategy is here.
IrishBoy123
  • IrishBoy123
compare like terms in each series
ganeshie8
  • ganeshie8
I 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
  • Hitaro9
Huh. 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
  • IrishBoy123
yeah, 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
  • anonymous
$$ \sum \frac{\frac 1 n }{1 + \frac 1 n }= \sum \frac {1}{n+1 }= \infty $$
ganeshie8
  • ganeshie8
If \(\lim\limits_{k\to\infty} a_k = 0\), 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
  • ganeshie8
looks im just repeating what irishboy has said
IrishBoy123
  • IrishBoy123
no, i chickened out of your second conclusion. can't remember why now :-)
Hitaro9
  • Hitaro9
So why does 1/1+ak -> that it diverges exactly? Sorry if this is obvious.
ganeshie8
  • ganeshie8
hmm question is not so clear... could you elaborate a bit
anonymous
  • anonymous
limit comparison test: http://prntscr.com/8j65pv
Hitaro9
  • Hitaro9
Oh wait right. Thank you Jayz. Okay then that makes sense.
Hitaro9
  • Hitaro9
So basically we have two cases, one for which ak -> 0 and one for otherwise? If I understand correctly?
anonymous
  • anonymous
:) no problem. ganeshie and irishboy did the hard work
anonymous
  • anonymous
oh you mean if a_k converged, then yes
Hitaro9
  • Hitaro9
Okay 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
  • anonymous
We 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
  • Hitaro9
Thanks all of you for the help. I appreciate it.
Hitaro9
  • Hitaro9
Alright, I gotta get to class, thanks again for all your help. You're all amazing people.
anonymous
  • anonymous
Just 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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