# P-Series or Comparison Test Question

1. Oct 31, 2014

### RJLiberator

K≥0 ∑ ((sqrt(k)+2)/(k+5))

I am trying to prove that this diverges. The divergence test is inconclusive.
Now I am left with a great option of a comparison test. I'm not quite sure what to compare it with, but I know I need to compare it with something smaller (denominator is larger) that diverges to prove divergence.

While looking at it, I am wondering if this is as simple as using the p-series test and taking off the smaller quantities of 2 and 5 and just using k^(1/2)/k where p would then = 1/2 and prove divergence.

Can I do this with the p-series?

IF not, I need to find something that is in comparison with this to prove divergence. Would k^(1/2)/2k be ok to use?

2. Oct 31, 2014

### vela

Staff Emeritus
This last line of reasoning is good. You care about what the terms do as k gets large, so looking at the dominant terms in the numerator and denominator is a good idea.

For large k, the series looks like $k^{-1/2}$, so using the limit comparison test with $k^{-1/2}$ might work.

You could also make an argument that the terms of the series are larger than a series of the form $c/\sqrt{k}$ and use the direct comparison test.

You can't say it's a p-series because it's not of the form $1/k^p$.

3. Oct 31, 2014

### RJLiberator

1) Ah, I cannot say it is a p-series because it is not in correct form. Great information.
2) Limit comparison test seems like a great idea, I may not be able to use it since we will be learning that later in the course.
3) I will need to use the direct comparison test that you pointed out to me. I am going to try this work now.

Thanks.

4. Oct 31, 2014

### RJLiberator

Since the sum starts at a value of 0 I cannot use k^(-1/2) alone in the denominator.

Can I compare it with c/(sqrt(k+1)) ?

5. Oct 31, 2014

### vela

Staff Emeritus
It would be simpler to throw away the first term. In other words, if the sum from k=1 converges, then the sum from k=0 will converge as well because the two series only differ by a finite number of terms.

The reason a series of the form $c/\sqrt{k}$ is nice to have is because it is a p-series, and you can therefore conclude it diverges.

6. Oct 31, 2014

### RJLiberator

Ah..... That makes sense. Being able to throw away initial 0 as it doesn't really add anything to the convergence/divergence of the sum. Thank you.

7. Oct 31, 2014

### Ray Vickson

You tell us. You need to have some confidence in your own knowledge/abilities.