Convergence between two sequences 2 If $a_{n}$ converges to $a$ and $b_{n}$ converges to $b$, then the sequence $<a_{n}b_{n}>$ converges to $ab$.
proof: Proof: Let $\epsilon > 0$, since $a_{n}$ converges to $a$, then there exists a positive integer $n_{0}$ such that $$n > n_{0} \Rightarrow |a_{n} - a| < \epsilon$$ Also, since $b_{n}$ converges to $b$ then there also exists a positive integer $n_{0}$ such that $$n > n_{0} \Rightarrow |b_{n} - b| < \epsilon$$. So, now we can choose an $N\in \mathbb{N}$ such that $|a_{n} - a| < \epsilon^{2}$ and $|b_{n} - b | < 1/\epsilon \ \  \forall n\geq N$. Hence, $\forall n\geq N$ we have $$|a_{n}b_{n}| < \epsilon^{2}\times 1/\epsilon = \epsilon$$. Therefore, $|a_{n}b_{n}|$ converges to $ab$.
Note: I am not sure if I am right with this proof but I don't see why it would be wrong, any suggestions would be greatly appreciated. 

Let's choose an example pair of sequences to see how your proof doesn't work. Suppose that $a_n = b_n = 1 + \frac{1}{n}$, both clearly going to $1$.
You mention that you can find sufficiently large $n$ such that $| a_n b_n | = (1 + \frac{1}{n})^2 < \epsilon$. But of course, that isn't true for small $\epsilon$, as $a_nb_n$ will always be larger than $1$ (and should go to $1$ anyhow!).
