Question

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

Answer 1

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?

Answer 2

\[\sum_{k=1}^{\infty}a _{k}\div(a _{k}+1)\]

Answer 3

Hey, last problem. Thanks for your help on the last one @IrishBoy123. I'm just not sure what the general strategy is here.

Answer 4

compare like terms in each series

Answer 5

I think you may simply use the limit test :

\(\lim\limits_{k\to\infty} \dfrac{a_k}{1+a_k} \ne 0 \implies \sum\limits_{k=0}^{\infty}\dfrac{a_k}{1+a_k} \text{ diverges}\)

Answer 6

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?

Answer 7

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

Answer 8

$$ \sum \frac{\frac 1 n }{1 + \frac 1 n }= \sum \frac {1}{n+1 }= \infty $$

Answer 9

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.

Answer 10

looks im just repeating what irishboy has said

Answer 11

no, i chickened out of your second conclusion. can't remember why now :-)

Answer 12

So why does 1/1+ak -> that it diverges exactly? Sorry if this is obvious.

Answer 13

hmm question is not so clear... could you elaborate a bit

Answer 14

limit comparison test:
http://prntscr.com/8j65pv

Answer 15

Oh wait right. Thank you Jayz. Okay then that makes sense.

Answer 16

So basically we have two cases, one for which ak -> 0 and one for otherwise? If I understand correctly?

Answer 17

:) no problem. ganeshie and irishboy did the hard work

Answer 18

oh you mean if a_k converged, then yes

Answer 19

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

Answer 20

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

Answer 21

Thanks all of you for the help. I appreciate it.

Answer 22

Alright, I gotta get to class, thanks again for all your help. You're all amazing people.

Answer 23

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