Proving Convergence of Sum: $\sum_{k=1}^{\infty}\frac{k+4}{k^3}$

[Saladsamurai]

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

 

[cristo]

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?

 

[mjsd]

use b_k = 1/k^2 instead

 

[Saladsamurai]

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?

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

use b_k = 1/k^2 instead

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 ?

Someone please elaborate.

Casey

 

[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

 

[Saladsamurai]

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

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?

Thanks for your patience,
Casey

 

[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..."

 

[Saladsamurai]

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..."

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

 

[mjsd]

this site has a few examples, it may help you visualise your concern
http://archives.math.utk.edu/visual.calculus/6/series.9/index.html

 

[cristo]

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.

 

[Saladsamurai]

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.

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

 

[cristo]

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

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.