Prove 2^(-n) converges

1. Feb 9, 2010

jrsweet

1. The problem statement, all variables and given/known data
Using the definition of convergence to prove that the sequence {2^(-n)} converges

2. Relevant equations

3. The attempt at a solution
So, I just don't think I am thinking straight or something. Here is what I got so far:

Chose e>0. Let N be any positive integer greater than ______.

How do I chose the N? Do I need to compare 2^(-n) to something larger to be able to find an N? Any help would be much appreciated. Thanks!

2. Feb 9, 2010

Staff: Mentor

First off, if you're going to prove that the sequence {2-n} converges to some value, you have to know what that value is. In other words, what is
$$\lim_{n \to \infty} 2^{-n}?$$

I don't see any evidence that you know what this value is; at least you didn't include that information in your post.

Second, show us that you know the definition of a sequence converging to some value.

3. Feb 9, 2010

jrsweet

Chose e>0. Let N be any positive integer greater than 1/e. Then, for n>=N we have

|1/(2n)-0| < |1/n|= (1/n) <= (1/N) < (1/(1/e)) = e

Thus, the sequence converges to 0??

Does that look right?

4. Feb 9, 2010

Dick

Yes, that looks ok. But you really didn't need the 1/n part. You could have just tried to find an N such that 1/2^N<e.

5. Feb 10, 2010

jrsweet

I was confused how to find it that way. I tried this:

1/2N < e
1/e < 2N
ln(1/e) < Nln2
ln(1/e)/ln2 < N

does that work?

6. Feb 10, 2010

Dick

Yes. Pick N>ln(1/e)/ln(2).

7. Oct 27, 2011

malachi31

I used the Ratio Test. I am not sure I did it correctly but here it goes.

lim n->inf {[2^(-n+1)]/[2^(-n)]}
= lim n-> inf {[2^(-n)*(1/2)]/[2^(-n)]
=lim n-> inf (1/2) = 1/2 < 1 Therefore by the ratio test the series converges.

Does this look right?

8. Oct 27, 2011

Dick

That's fine. But the original problem was to show the SEQUENCE converge by the DEFINITION (using epsilons etc). You proved the SERIES converges using the RATIO test. That does imply the sequence converges, but it's not what's being asked for.