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. 

Hint: The identity you might want to use is $|a_nb_n-ab| = |b_n(a_n-a) + a(b_n-b)| = |b_n|\cdot|a_n-a| + |a|\cdot|b_n-b|$. Next use the fact that $b_n\stackrel{n\rightarrow\infty}{\rightarrow}b$ to say that the sequence $(b_n)$ must be bounded.
