Three questions, hope that's not too many!

1. I need to prove that the sequence

$\displaystyle a_n = 1 + \frac{1}{2} + \frac{1}{3} +...+\frac{1}{n}$

What I have is for any $\displaystyle \varepsilon$ there exists $\displaystyle m,n \geq n*$ such that $\displaystyle |a_m - a_n|$

I take that as $\displaystyle |(1 + \frac{1}{2} +...+\frac{1}{m}) - (1 + \frac{1}{2} +...+\frac{1}{n})| $

and get

$\displaystyle |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})| $

Here is my first question: when I put that into

$\displaystyle |\frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{m})| > \frac{1}{n}$

I need a fixed vaule for $\displaystyle \varepsilon$. 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 $\displaystyle a_n$ converges to 0, and a sequence $\displaystyle b_n$ is bounded, then the sequence $\displaystyle a_n b_n$ converges to 0.

I had gotten help on this before, but hadn't gotten very far. I'm starting with

$\displaystyle \left[ {\exists B > 0} \right]\left( {\left| {b_n } \right| < B} \right)\quad \& \quad \varepsilon = \frac{\delta }{B}$, but how would I put this into a proof for this? again, confused!

Thanks in advance, and I might have some more later!