# Thread: Using Limit Comparison Test/Basic Comparison Test on series

1. ## Using Limit Comparison Test/Basic Comparison Test on series

Hi all,

So I have a homework assignment that I was hoping to have clarified.

The first problem is:

$\displaystyle \sum_{k=8}^{\infty} \frac{4}{2+4k}$ (by the basic comparison test)

My idea is that $\displaystyle 0 \leq \frac{4}{2+4k} \leq \frac{4}{4k}$ which I assume you factor out the 4 and make it the harmonic series. If anyone could help on that one to either confirm that one or point me in the right direction that'd be awesome.

The second problem is:

$\displaystyle \sum_{k=6}^{\infty} \frac{k^3+2k+3}{k^4+2k^2+4}$ (by the limit comparison test)

I found it to be $\displaystyle \sum_{k=6}^{\infty}\frac{1}{k}$ by simply taking $\displaystyle \frac {k^3}{k^4}$ which is the harmonic series $\displaystyle \frac{1}{k}$.

However, I don't know if the tail end theorem would apply and it would diverge, or to compensate for the k=6, I would have to make series $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^6}$ which is a p-series that converges.

So, being stuck between those two (although I believe it's harmonic and diverges), any help is very much appreciated!

Thanks to anyone who does help with one (or both) of the problems!

2. ## Re: Using Limit Comparison Test/Basic Comparison Test on series

For the first problem the series diverges so find a divergent series which has smaller terms.

For the second problem,
If the series you are given is $\displaystyle \sum a_n$ then find a series $\displaystyle \sum b_n$
So that the sequence $\displaystyle \frac{a_n}{b_n}$ converges, then both series converge or both series diverge.

3. ## Re: Using Limit Comparison Test/Basic Comparison Test on series

So for question one, would a way to do that be the opposite of what I did?

$\displaystyle \frac{4}{2+4k} \leq \frac{1}{2+4k} \leq \frac{1}{4k}$

Also for number two, I wasn't necessarily looking for how to use the LCT, just looking to see if the series was harmonic by the tail end theorem or not.

4. ## Re: Using Limit Comparison Test/Basic Comparison Test on series

Hi billb91

For Question 1, your assumption was correct.
compare with 4/(4k) and 4/(4k) diverge by harmonic series so the smaller series diverges as well.

For Question 2, you LCT was right too.
comparing with 1/k is good.
The limit should give you a number so the series diverge, since the one that you are comparing with is 1/k which is harmonic series (diverges).

You can take a look at some summary that I posted
eBooks | Tutorat A+ Tutoring

Hope this help
Francis
Tutorat A+ Tutoring | Go for A+ , A big step toward the future.