simple convergence property proof (Sequences) Question: Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences with $a_n \Rightarrow L$ and $b_n \Rightarrow M$ as $n \Rightarrow \infty$. 
Prove that $a_nb_n \Rightarrow LM$
Solution: (My Attempt). Instead of redoing it could someone just tell me what I'm doing wrong. Thx
WTS: 
(1) $\exists L \in R, \forall \epsilon > 0, \exists N_1 > 0$ such that for all $n \in N_1$, if $n > N_1$, then 
$|a_n - L| < \text{(We dont know yet)}$
(2) $\exists M \in R, \forall \epsilon > 0, \exists M > 0$, such that for all $m \in M$, if $m > M$, then 
$|b_n - M| < \text{(We dont know yet)}$
Choose N = $\text{we dont know yet} > 0$
Suppose $n > N$ and $m > M$, then
$$|a_nb_n - LM| = |a_nb_n - a_nM + a_nM - LM| $$
$$= |a_n(b_n - M) + M(a_n - L)| \text{ by algebra}$$
$$\leq |a_n(b_n-M)| + |M(a_n - L)| \text{ triangle inequality}$$ 
$$= |a_n||b_n - M| + |M||a_n - L|$$
Can we say $|a_n||b_n - M| = \epsilon/2$ same with $|M||a_n - L| = \epsilon/2$ ? Then Q.E.D? With N = $max(N_1, M)$ ? 
I have no idea what I'm doing. 
 A: You know that there exists $\overline{N_1}$ such that for every $n>\overline {N_1}$ you have $|L|-1 \leq |a_n| \leq |L|+1$, and similarly there exists $\overline{N_2}$ such that for every $n> \overline {N_2}$ you have $|M|-1 \leq |b_n| \leq |M|+1$.
Then the terms $|a_n|$ and $|b_n|$ are bounded and you just have to pick an  $\epsilon>0$ according to them.
A: you messed up the notations were you assumed $b_{n}$ converges to M and in the proof you for all $m > M$ .
So I am redoing the same proof  
I am assuming both the sequences are real 
In order to prove the given statement it is helpful to note that every real convergent sequence is a cauchy sequence and hence a bounded sequence.
Let $\{a_{n}\}_{n\geq 1}$ be a sequence of real numbers converging to $a$ and $\{b_{n}\}_{n\geq 1}$ be a sequence of real numbers converging to $b$.
$|a_{n}b_{n} - ab| = |a_{n}b_{n} - a_{n}b+a_{n}b-ab|= |a_{n}(b_{n}-b)+b(a_{n}-a)|$
therefore 
$|a_{n}b_{n} - ab|\leq |a_{n}||b_{n}-b|+|b||a_{n}-a|$  for all $n\in \mathbb{N}$
As noted above $a_{n}$ is a bounded sequence, so there exists a real number $\alpha$ greater than zero such that $|a_{n}| \leq \alpha$ for all n.
Therefore  
$|a_{n}b_{n} - ab|\leq \alpha |b_{n}-b|+|b||a_{n}-a|$  for all $n\in \mathbb{N}$
now choose $\epsilon_{1}=\frac{\epsilon}{\alpha}$ for some $\epsilon > 0$ and if $b \neq 0$ then choose $\epsilon_{2}=\frac{\epsilon}{|b|}$.
So for $\epsilon_{1}$ there exists a $N_{1} \in \mathbb{N}$ such that 
for all $n\geq N_{1}$ $|b_{n}-b|<\epsilon_{1}$ -------(*)
So for $\epsilon_{2}$ there exists a $N_{2} \in \mathbb{N}$ such that 
for all $n\geq N_{2}$ $|a_{n}-a|<\epsilon_{2}$ -------(**)
Let N = max{$N_{1},N_{2}$} 
then by (*) and (**)
$|a_{n}b_{n} - ab|\leq \alpha |b_{n}-b|+|b||a_{n}-a| < \alpha \epsilon_{1}+|b|\epsilon_{2} $ for all $n\geq N$. 
$|a_{n}b_{n} - ab| < 2\epsilon $ for all $n\geq N$.
Thus $a_{n}b_{n}$ converges to $ab$.
