Series Help - convergence: What values of k is the sequence {k^n) n -> convergent?

[wax4brains]

For what values of k is the sequence {k^n} convergent?

 

[HallsofIvy]

What values of k is the sequence {k^n) n -> convergent?
I have the answers of -1 and 1 but I do not understand it

It's hard to know how to answer if we do not know what you do understand! I hope you are not saying that you do not know how to multiply! Do you know what a "sequence" is? Do you know what it means for a sequence to converge?

If k= 1, k^2= (1)(1)= 1, k^3= 1(1)= 1, .... The sequence is 1, 1, 1, ... and "converges" to 1. If k= -1, k^2= (-1)(-1)= 1, k^3= (-1)(1)= -1, k^4= (-1)(-1)= 1.... The sequence is -1, 1, -1, 1, -1, 1, ... and does NOT converge.

You know, don't you, that if a> 1 that ab> b but if a< 1 ab< b (a and b positive). So if k> 1, k^2 is larger, k^3 is even larger, k^4 larger yet. The sequence does not converge.

If -1< k< 1, k^2 is closer to 0, k^3 even closer, etc. For any k between -1 and 1, the sequence converges to 0.

The sequence k^n converges for -1< k<= 1 and does not converge for any other values of

 

[Dr.Peterson]

What values of k is the sequence {k^n) n -> convergent?
I have the answers of -1 and 1 but I do not understand it

Just to make sure we understand your question, here is how I am modifying it in order to make sense of it: You have mismatched brackets (a brace and a parenthesis), and the notation "n -> convergent" is not standard. I think either you meant

For what values of k is the sequence {k^n} convergent, as n->infinity?
​

or possibly

For what values of k is the sequence {(k^n)n} convergent?
​

with the arrow just meaning something about an implication.

The first, for k=2, would be 2^1, 2^2, 2^3, ..., i.e. 2, 4, 8, ...; the second would be 2^1*1, 2^2*2, 2^3*3, ..., i.e. 2, 8, 24, ... .

Please tell us what you did mean, and why you think 1 and -1 are the answers, and what you do know about convergence of a sequence.

I wonder if the solution you saw was (-1, 1], in interval notation, and you took that to be the set {-1, 1}?

 

[Steven G]

What values of k is the sequence {k^n) n -> convergent?
I have the answers of -1 and 1 but I do not understand it

I do not agree with your answer at all.

Let's look at when k=1. Then k^n = 1^n = 1 for all n. So if k=1, (k^n)n will simply be (1)n or n. But n goes without bound as n-->oo, as a result of this the sequence diverges when k=1

Let's now assume that k= -1. (-1)^n will oscillate between -1 and 1. So if k= -1, (k^n)n will oscillate between -n and n. As n gets larger, -n and n grow further and further apart and hence will not converge either.

 

[Dr.Peterson]

I do not agree with your answer at all.

Let's look at when k=1. Then k^n = 1^n = 1 for all n. So if k=1, (k^n)n will simply be (1)n or n. But n goes without bound as n-->oo, as a result of this the sequence diverges when k=1

Let's now assume that k= -1. (-1)^n will oscillate between -1 and 1. So if k= -1, (k^n)n will oscillate between -n and n. As n gets larger, -n and n grow further and further apart and hence will not converge either.

This is one reason I think my first guess as to what he meant, which HallsofIvy assumed, is the correct one, not what you are supposing.

But we'll have to wait to find out.