# Thread: comparison test with 0 as upperbound

1. ## comparison test with 0 as upperbound

We all know the rules for comparison test with 0 <= asubk <= bsubk. Then

I. If my series of a's diverges then so does my series of b's.
II. if my series of b's converges then so does my series of a's.

However, I want to know what these two rules would look like if

asubk <= bsubk <= 0. I know the test still works.

I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

Thanks

2. Originally Posted by grandunification
asubk <= bsubk <= 0. I know the test still works.

I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

Thanks
if the a's, then the b's will. if the b's diverge, then the a's will.

try to modify the original comparison test by multiplying the system through by a negative power. if 0 is a lower bound, that is.

it would do you well to actually post the problem you are having trouble with

3. Instead of the series of a's, try to use the series of (-a)'s, and so is for b

4. Originally Posted by grandunification
We all know the rules for comparison test with 0 <= asubk <= bsubk. Then

I. If my series of a's diverges then so does my series of b's.
II. if my series of b's converges then so does my series of a's.

However, I want to know what these two rules would look like if

asubk <= bsubk <= 0. I know the test still works.

I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

Thanks
A series of non-negative terms converges if and only the sequence of partial sums are bounded. Therefore, if $\displaystyle \sum_{n\geq 1}b_n$ converges then $\displaystyle \sum_{k=1}^n b_k$ is bounded, but so is $\displaystyle \sum_{k=1}^n a_k$, and so $\displaystyle \sum_{n\geq 1}a_n$ converges.