# Sequences and Series

1. Apr 7, 2014

### cathy

1. The problem statement, all variables and given/known data

So, I actually have a bunch of these problems and I cannot do any of them. I don't think I'm really understanding it. Here is the question: (one of them) The way I wrote them, a_n means a sub n

For each sequence a_n find a number k such that n^k a_n
has a finite non-zero limit.
(This is of use, because by the limit comparison test the series ∑from 1 to infinity (a_n) and ∑from 1 to infinity (n^-k) both converge or both diverge.)

The question:
a_n= (6+3n)^-7
What does k equal?

2. Relevant equations
N/A

3. The attempt at a solution
I'm actually 100% lost on this problem. I don't even know where to start :(
Here's my attempt. It may not make any sense because I don't understand this. But here it is:
(6+3n)^-7>0
1/(6+3n)^7>0
1>0
Ans: 1
But that is incorrect.

2. Apr 7, 2014

### Dick

Here's a hint. (6+3n)=n(6/n+3). Try and do something with that.

3. Apr 7, 2014

### cathy

Would you mind explaining the meaning? How would I find k from it?

4. Apr 7, 2014

### Dick

Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). What's the limit n->infinity of (6/n+3)^(-7)?

5. Apr 7, 2014

### cathy

isn't it infinity?

6. Apr 7, 2014

### Dick

I don't think so. Why do you? What's limit 6/n as n->infinity?

7. Apr 7, 2014

### cathy

because isnt it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?

8. Apr 7, 2014

### cathy

but how would doing that help me find what k is?

9. Apr 7, 2014

### Dick

No, it isn't. ((6/n)+3)^(-7)=1/((6/n)+3)^7. It's not 6/(n+3). It's (6/n)+3. They are very different.

10. Apr 7, 2014

### cathy

and using this new limit (6/n)+3, how would it lead me to solve for k?

11. Apr 7, 2014

### Dick

I think you should stop asking that until you do the algebra correctly and tell me what limit n->infinity ((6/n)+3) is. Then take a breath and think about it.

12. Apr 7, 2014

### cathy

the limit is 3 because the 6/n goes to 0

13. Apr 7, 2014

### Dick

Ok, so what's limit ((6/n)+3)^(-7). Have you thought about what that might mean for your question of what k is?

14. Apr 7, 2014

### cathy

1/3^7= 1/2187.
I still dont know where k comes into play.

15. Apr 7, 2014

### cathy

Is k 7?

16. Apr 7, 2014

### cathy

Where did you get the n(6/n+3) here? Why was it necessary to factor out the n?

17. Apr 7, 2014

### Dick

Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). As n->infinity (6/n+3)^(-7)->1/3^7. Does it make sense to you that k=7 works??

18. Apr 7, 2014

### Dick

Because you want the series to look like n^(-k) times something that approaches a nonzero constant. It's a good idea to factor out the n and see what's left.

19. Apr 7, 2014

### cathy

yes it does. so just to recap, i had to factor out an n to create the n^k? what if n were in the demonimator instead?

20. Apr 7, 2014

### Dick

Then you would factor out a 1/n^(k)=n^(-k). It just changes the sign of k. Doesn't it?