Hard infinite series converges problem (Real Analysis)

[nevnight13]

Homework Statement

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

Homework Equations

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

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.

 

[Dick]

Homework Statement

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

Homework Equations

From the Given statements, we know 1+bk<1 and 0<bk/(1+bk)<1

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.

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

 

[jbunniii]

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

 

[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.

 

[Seydlitz]

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.

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 $b_k \geq 1$?

 

[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.

 

[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.

 

[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.