Finding a convergent subsequence of the given sequence

[Maddiefayee]

Homework Statement

For some background, this is an advanced calculus 1 course. This was an assignment from a quiz back early in the semester. Any hints or a similar problem to guide me through this is greatly appreciated! Here is the problem:
Find a convergent subsequence of the sequence:
{(-1)ⁿ (1-(1/n)}^∞_n=1

Homework Equations

I don't think there are any equations needed. The class is all about proofs. Here's a definition:
A sequence {a_n} is said to converge to the number a provided that for every positive number ε there is an index N such that:
|a_n - a| < ε , for all indices of n ≥ N

The Attempt at a Solution

So this was my attempt. My "solution" was: {(1/n²)(-1)ⁿ}¹⁰_n=2
How I got to this solution was honestly listing out a few terms of the original sequence and then finding another sequence that I thought would make sense.

Here's the note from my professor: "Sequences, and subsequences, have an infinite number of terms. Also, this sequence is not a subsequence."

 

[haruspex]

Here's the note from my professor:

So do you now understand what constitutes a subsequence and why your attempt was not one?

 

[Maddiefayee]

So do you now understand what constitutes a subsequence and why your attempt was not one?

Yes! I understand that it is not a subsequence. I think I am just overthinking the problem here.

 

[haruspex]

Yes! I understand that it is not a subsequence.

Good.

So, write out four or five terms of the full sequence. What is it about the full sequence that means it will not converge?

 

[Maddiefayee]

It won't converge because it oscillates between positive when n is even and negative when n is odd.

 

[haruspex]

It won't converge because it oscillates between positive when n is even and negative when n is odd.

Well, that's not enough in itself to prevent convergence. (-1)ⁿ/n converges happily. But you are right that it is part of the problem here. So how can you select a subsequence to avoid it?

 

[Maddiefayee]

I could make it so that n=2k where k is in ℝ.

 

[haruspex]

where k is in ℝ.

You mean ℕ, right?
What would that give? Can you show it converges? (Or maybe you are not required to prove it.)

 

[Maddiefayee]

Right. So then the new sequence would be:

{(-1)²ⁿ(1-(1/2n))}^∞_n=1

It would converge to 1.

 

[haruspex]

Right. So then the new sequence would be:

{(-1)²ⁿ(1-(1/2n))}^∞_n=1

It would converge to 1.

Right, but please use parentheses correctly: 1-1/(2n).