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

1. 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?

2. hitaro9

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

3. hitaro9

|dw:1442920394346:dw|

4. hitaro9

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

5. IrishBoy123

compare like terms in each series

6. ganeshie8

I think you may simply use the limit test : $$\lim\limits_{k\to\infty} \dfrac{a_k}{1+a_k} \ne 0 \implies \sum\limits_{k=0}^{\infty}\dfrac{a_k}{1+a_k} \text{ diverges}$$

7. 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?

8. 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

9. anonymous

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

10. 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.

11. ganeshie8

looks im just repeating what irishboy has said

12. IrishBoy123

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

13. hitaro9

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

14. ganeshie8

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

15. anonymous

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

16. hitaro9

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

17. hitaro9

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

18. anonymous

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

19. anonymous

oh you mean if a_k converged, then yes

20. 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

21. 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

22. hitaro9

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

23. hitaro9

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

24. 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