# Homework Help: Showing Convergence

1. Dec 9, 2007

Show that $$\sum_{k=1}^{\infty}\frac{k+4}{k^3}$$ converges.

I thought I would try the limit comparison test by using $$a_k=\frac{k+4}{k^3}$$ and $$b_k=\frac{1}{k^3}$$.

I thought a_k looked similar to power series so that if the lim ask-->infinity of a_k/b_k is finite and > 0 then since the p-series (for p>1) converges, than so must a_k,

But I am getting an infinite limit.

So my questions are:

1.) Where is my reasoning flawed?

2.) What is the correct approach?

~Casey

2. Dec 9, 2007

### cristo

Staff Emeritus
Your original series can be written $$\sum_{k=1}^\infty\frac{1}{k^2}+\frac{4}{k^3}$$. Does this give you a hint as how to proceed?

3. Dec 9, 2007

### mjsd

use $$b_k = 1/k^2$$ instead

4. Dec 9, 2007

I like this, but we shouldn't need to use partials.

I see that this will give me a limit of 1...but you have not given me any justification as to WHY I should or even can use p=2 as opposed to p=3 when the original series had p=3 ?

Casey

5. Dec 9, 2007

### mjsd

Read the limit comparsion test description again....where does it say that the p-series used must be related to the original series in any way?

EDIT: all you require is that both a_k and b_k are positive for all k

6. Dec 9, 2007

Well, my instructor is forever implying that we should use a series that relates. But you are correct, the description does not say that it needs to.

That being said, why can I not use p= anything, i.e, 3 in this case? I s can use anything...why NOT 3?

Casey

7. Dec 9, 2007

### mjsd

you aim here is to find a series b_k that will help you show whether sum of a_k is convergent or not. Note that a_k ~ 1/k^2 when k is large, that's why using b_k = 1/k^2 will help because we know that a_k/b_k ~ 1/k^2 / (1/k^2) ~ finite. or at least the first thing one would try

yeah.. in a sense, that's why you should use a series that relates.... but it is more subtle than just: "since that you have a k^3 on the bottom for a_k so it would means p=3..."

8. Dec 9, 2007

I am still not clear on why if all we need to do is use a b_k such that we know the outcome of it....and we know $$\sum b_k=\sum_1^{\infty}\frac{1}{k^3}$$ converges and we also know that
$$\sum a_k=\sum_1^{\infty}\frac{k+4}{k^3}$$ converges...then why does there the limit at infty of a_k/b_k not yield a finite number...

I mean I know that it doesn't because the limit is what it is....but shouldn't my selection of b_k, by the definition of the limit comparison test, be completely arbitrary?

Thanks again,
Casey

Last edited: Dec 9, 2007
9. Dec 9, 2007

### mjsd

10. Dec 9, 2007

### cristo

Staff Emeritus
The point is that you're looking at the test the wrong way around. The limit comparison test says that if $\lim_{k\to\infty}\frac{a_k}{b_k}>0$ , then a_k converges iff b_k converges. If the limit is infinite, and b_k converges, it does not tell you anything about a_k.

Last edited: Dec 9, 2007
11. Dec 9, 2007

So if I know the outcome of b_k but the lim of a_k/b_k is infinite...then I need to make another comparision since nothing can be learned from this.

Does that sound correct?

Casey

12. Dec 9, 2007

### cristo

Staff Emeritus
If the limit is infinite and b_k converges, then you need another test. If the limit is infinite and b_k diverges, then as stated on mjsd's link above, a_k also diverges.