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. 
 A: 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.
A: 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!).
A: The relation you are using with $|a_n−a|<ϵ^2$ and $|b_n−b|<1/ϵ$ $\forall n\geq N$ is $|a_nb_n|< |a_n-a|\times |b_n-b|$ 
which may not hold always. The inequality you can use is
$|a_nb_n-ab| = |(a_n-a)(b_n-b) + a(b_n-b) + (a_n-a) b | $
\begin{align*}
|a_nb_n - ab|&\leq |(a_n-a)(b_n-b)| + |a||(b_n-b)| + |(a_n-a)| |b| \\
&\leq \epsilon^2 +(|a| + |b|) \epsilon
\end{align*}
A: If you are looking for an credible/official source, this exact statement can be found in the book "Analysis I" by Amann & Escher. It's the English version of a standard undergraduate introductory textbook to Analysis. You can find it on Amazon easily.
In the book, you can find it on page 142 as Proposition 2.4 (ii):
Let $(x_n)$ and $(y_n)$ be sequences in $\mathbb{K}$ (where $\mathbb{K} = \mathbb{R}$ or $\mathbb{C})$. 
If $\lim x_n = a$ and $\lim y_n = b,$ then $\lim(x_n y_n) = ab.$
Proof: Since $x_n \rightarrow a, (x_n - a)$ is a null sequence. Since any convergent sequence is bounded (which is Proposition 1.10 in the book, page 137), $(y_n)$ is bounded. Then $((x_n - a)y_n)_{n \in \mathbb{N}}$ is a null sequence. Since $(a(y_n - b))_{n \in \mathbb{N}}$ is also a null sequence, we have:
$x_n y_n -ab = (x_n -a)y_n + a(y_n - b) \rightarrow 0 \quad (n \rightarrow \infty).$
So the sequence $(x_ny_n)$ converges to $ab$.
