# Direct comparison test, just need some explanation

1. Jul 29, 2011

### shemer77

2 problems, I need to use the direct comparison test with either a p series or a geometric series
1)series of j^2/(j^3 +4j +3)
I thought of comparing it to j^2/J^3 which comes out to 1/j, but that dosent work, my teachers answer is you compare it to 1/5j
2) series of sqrt(q)/(q+2) I would figure it would be 1/q^(1/2), but nope that dosent work, my teachers answer is comapre it to 1/3q^(1/2)

Can you explain to me how he got those? i feel like all he did was multiply the denominator by some arbitrary number to get an answer that suited his needs? is there more to this?

2. Jul 29, 2011

### gat0man

1.

In order to satisfy the conditions of the comparison test, if you want to show your original series diverges, you need something smaller or equal to. You can *guarantee* that the series you are comparing it to will be smaller by taking something with a larger coefficient than the 4j, such as.... 5j. Basically if you chose 1/j, the series would *not* be smaller for all n>0. So by choosing 5j, you satisfy the conditions of the comparison test.

2. Again, this is an issue of thinking "If I compare it to series b, will series b always be smaller than series a?" The series you are using to compare it to will *not* be smaller unless you take your series to be at least 1/3j^(1/2)

What I like to do is when you are faced with something simple along these lines, test a few values of n such as n = 1, n = 2, and see if the series you are trying to use will meet the requirements for the comparison test. For instance in 2. if you take n=1, your comparison b would be 1, but the original series would be 1/3, so your series would *not* be smaller. Since you need it to be equal to or smaller, make your comparison series something that will be at least 1/3 at n=1 and still meet the requirements of the test.

Edit: I suppose it would be somewhat arbitrary to meet your needs of the test, but that is the beauty of the comparison test. You could have chosen for the first one 1/(100000 * j) and still have been correct in it's usage.

Last edited: Jul 29, 2011
3. Jul 29, 2011

### Dick

Yes, there's a little more to it. It's true j^2/(j^3+4j+3) is basically like 1/j. But you know the sum 1/j DIVERGES. To prove j^2/(j^3+4j+3) diverges you need a comparison series that diverges that is LESS than j^2/(j^3+4j+3). The problem with using 1/j is that 1/j is GREATER than j^2/(j^3+4j+3). Showing a series is less than a divergent series is no help. That's what the extra factors are about.

4. Jul 30, 2011

### Bohrok

If you start with 4j3 > 4j + 3, this implies 5j3 > j3 + 4j + 3 (for j > 2).
Can you see how using this comparison really just comes down to what your teacher said with comparing to 1/5j?