Three questions, hope that's not too many!
1. I need to prove that the sequence
What I have is for any there exists such that
I take that as
and get
Here is my first question: when I put that into
I need a fixed vaule for . What am I missing?
2.
Prove that every unbounded sequence contains a monotone subsequence that diverges to infinity.
On this one, I know unbounded can mean divergent. What do I do next to prove this? I am confused on this one!
3.
Prove that if a sequence converges to 0, and a sequence is bounded, then the sequence converges to 0.
I had gotten help on this before, but hadn't gotten very far. I'm starting with
, but how would I put this into a proof for this? again, confused!
Thanks in advance, and I might have some more later!
sorry, I forgot to finish the instructions. You are correct, I need to prove it it cauchy. I didn't look at it as a harmonic series, that seems to be the solution. would it be appropiate to just state that the harmonic series diverges or do I need to go farther into it?
Because the sequence u(n) is unbounded (I'll assume upwards without loss of generality), you can choose any M and find an index m so that u(m) > M. Choose an M with such a u(m) and take u(m) as first element of your subsequence. Since the sequence was unbounded, there will be an index p such that u(p) > u(m) (else, the sequence would have been bounded by u(m)). Take this one as second element so that you have a monotone sequence. Proceed with this argument, you are now constructing a monotone, divergent sequence.