# Hard infinite series converges problem (Real Analysis)

1. Dec 13, 2013

### nevnight13

1. The problem statement, all variables and given/known data
let bk>0 be real numbers such that Ʃ bk diverges. Show that the series Ʃ bk/(1+bk) diverges as well.
both series start at k=1

2. Relevant equations
From the Given statements, we know 1+bk>1 and 0<bk/(1+bk)<1

3. The attempt at a solution
I've tried using comparison test but cannot find anything that's definitely less than bk/(1+bk). Iv'e tried the ratio test but i get stuck with a bunch of bk+1's that I cannot do anything with.

Some help or at least a push in the right direction would be much appreciated. Thank you. I'm also kinda in a rush haha.

Last edited: Dec 13, 2013
2. Dec 13, 2013

### Dick

You must be in a rush. If bk>0 then 1+bk>1, not 1+bk<1.

3. Dec 13, 2013

### jbunniii

Hint: try considering separately the cases $b_k < 1$ and $b_k \geq 1$.

4. Dec 14, 2013

### nevnight13

Sorry but is there anymore guidance you could give? I keep failing at comparison and ratio test doesn't seem to work because we do not know if bk is decreasing, increasing or oscillating.

5. Dec 14, 2013

### Seydlitz

I'm not an expert and might be wrong, but if you could perhaps derive from the given hint, if $0<b_k<1$, then what happens with the other series? Does it pass the preliminary test? What happens to the other series if $b_k \geq 1$?

6. Dec 14, 2013

### jbunniii

Well, if $b_k < 1$ then $1 + b_k < 2$.

And if $b_k \geq 1$ then $2b_k \geq b_k + 1$.

Try using these facts to get a lower bound for the terms of your series in each case.

7. Dec 14, 2013

### nevnight13

Here is my attempt...:

case 1: bk<1 then 1+bk<2

so bk/(1+bk) > bk/2 so and Ʃ[bk/2] diverges so by comparison Ʃ[bk/(1+bk)] diverges

case 2: bk>=1 then 2bk>bk+1

so Ʃ[bk/2bk] < Ʃ[bk/(1+bk)]
-->Ʃ[1/2] < Ʃ[bk/(1+bk)]
since Ʃ[1/2] diverges Ʃ[bk/(1+bk)] diverges by comparison

I apologize for the bad format.

8. Dec 14, 2013

### jbunniii

OK, you have the right idea. Your solution is fine in case $b_k < 1$ for all $k$, or in case $b_k \geq 1$ for all $k$. But in general, it may be true that $b_k < 1$ for some $k$ and $b_k \geq 1$ for other $k$. Your proof needs to handle this general situation as well.