Show that if $\{x_n\}$ converge and $\{y_n\}$ diverge then $\{ax_n + by_n\}$ diverge, for $b \ne 0$ 
Let $\{x_n\}$ be a convergent sequence and $\{y_n\}$ a divergent sequence. Prove that $\{ax_n + by_n\}$ diverges for $b\ne 0$.

Intuitively this is obvious to me, however i want a formal proof.
Using the definition of a limit we know that:
$$
\lim_{n\to \infty}ax_n = A \stackrel{\text{def}}{\iff} \{\forall \varepsilon > 0\ \exists N \in \mathbb N : \forall n \ge N \implies |ax_n - A| < \varepsilon\}
$$
On the other hand we know that $y_n$ diverges, thus:
$$
\lim_{n \to \infty}by_n = \exists! \stackrel{\text{def}}{\iff} \{\exists \varepsilon >0 \ N \in \mathbb N : \exists n \ge N \implies |by_n - B| \ge \varepsilon\}
$$
Choose some $\varepsilon$:
$$
\begin{cases}
|ax_n - A| < {\varepsilon \over 2} \\
|by_n - B| \ge {\varepsilon \over 2}
\end{cases}
$$
or:
$$
\begin{cases}
|ax_n - A| < {\varepsilon \over 2} \\
-|by_n - B| \le -{\varepsilon \over 2}
\end{cases}
$$
This is where I got stuck. It feels like i have to use some sort of triangular inequality (either direct or reverse) and find a contradiction. How do I proceed from this point?
 A: There's more than one way to skin a cat.
The easiest way to prove your point is by knowing another fact, that is:

If $\{a_n\}$ converges and $\{b_n\}$ converges and $\alpha, \beta\in\mathbb R$, then $\{\alpha a_n + \beta b_n\}$ also converges.

Using this, you can easily construct a proof by contradiction. That is, you can, assuming that $\{x_n\}$ converges and $\{ax_n + by_n\}$ converges, prove that $\{y_n\}$ must also converge, since $$y_n = \frac{1}{b}\left(ax_n + by_n\right) + \left(-\frac{a}{b}\right)x_n$$

However, if you insist on going by definitions, then you shouldn't just "choose some $\epsilon$." The fact that $\{y_n\}$ does not converge gives you a place where you can start to build your $\epsilon$. In particular, choosing some $B$, you can take the $\epsilon_y$ which satisfies the property that $\forall N\in\mathbb N\exists n>N: |y_n - B|>\epsilon$.
You can then use this to prove that $a\cdot A + b\cdot B$ cannot be a limit, and since every number can be written as $aA+bB$ for some value $B$, this proves the sequence doesn't converge.
A: For the sake of completeness I'm putting here my further steps based on the hints.
Let:
$$
\lim_{n\to \infty}x_n = A\\
\lim_{n\to \infty}y_n = B\\
$$
We want to show that $\lim_{n\to\infty}(ax_n + by_n)$ exists. Suppose $x_n$ and $y_n$ converge. It is necessary and sufficient $(1)$ to show that:
$$
x_n = A + \alpha_n\\
y_n = B + \beta_n
$$
Where $\alpha_n$ and $\beta_n$ are infinitely small sequences. Then by definition of a limit we have:
$$
|x_n - A| < \varepsilon \iff |\alpha_n| < \varepsilon \iff \lim_{n\to\infty}\alpha_n = 0 \\
|y_n - B| < \varepsilon \iff |\beta_n| < \varepsilon \iff \lim_{n\to\infty}\beta_n = 0
$$
Take some $a$ and $b$ and consider the following limit:
$$
\lim_{n\to\infty}(ax_n + by_n) = \lim_{n\to\infty}\left((aA + bB) + (a\alpha_n + b\beta_n)\right) = aA + bB + \lim_{n\to\infty}(a\alpha_n + b\beta_n)
$$
But any linear combination of infinitely small sequences is an infinitely small sequence. So:
$$
\lim_{n\to\infty}(ax_n + by_n) = aA + bB
$$
Now consider the case when $y_n$ diverges, that means that:
$$
|y_n - B| \ge \varepsilon \iff |\beta_n| \ge \varepsilon \iff \lim_{n\to\infty}\beta_n = \exists !
$$
So since $y_n$ may not be presented as a sum of some constant and and infinitely small sequence (which violates $(1)$) then $(ax_n + by_n)$ must also diverge.
