Solve Sequences & Series Homework: Find k for (6+3n)^-7

[cathy]

Homework Statement

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?

Homework Equations

N/A

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.

 

[Dick]

Homework Statement

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?

Homework Equations

N/A

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.

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

 

[cathy]

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

 

[Dick]

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

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)?

 

[cathy]

isn't it infinity?

 

[Dick]

isn't it infinity?

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

 

[cathy]

because isn't it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?

 

[cathy]

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

 

[Dick]

because isn't it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?

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.

 

[cathy]

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

 

[Dick]

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

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.

 

[cathy]

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

 

[Dick]

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

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

 

[cathy]

1/3^7= 1/2187.
I still don't know where k comes into play.

 

[cathy]

Is k 7?

 

[cathy]

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

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

 

[Dick]

Is k 7?

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??

 

[Dick]

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

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.

 

[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?

 

[Dick]

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?

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

 

[cathy]

i have a problem that says 7/(n^3+n) you don't have to solve this, but i just want to know, would i have to factor out n? and i would have 7n^-3?

 

[Dick]

i have a problem that says 7/(n^3+n) you don't have to solve this, but i just want to know, would i have to factor out n? and i would have 7n^-3?

That will work. What's the limit of what's left after you factor that out?

 

[cathy]

is it 0?

 

[Dick]

is it 0?

Is that a guess? What's left over after you factor 7n^(-3)) out? Mind your algebra.

 

[cathy]

1/n^4

 

[Dick]

1/n^4

(7/n^3)*(1/n^4) is not equal to 7/(n^3+n). Do algebra, ok?