Proving the theorem about real series with positive members I would like to receive some help about the next problem:
The problem
I am trying to understand the proof of the following theorem about the two real series with positive members:
"Let it exist $\lim_{n \to +\infty} \frac{a_n}{b_n} = K$, $0 \le K \le +\infty$, where $a_n$ and $b_n$ are members of the real series with positive numbers $\sum_{n = 0}^{+\infty} a_n$ and $\sum_{n = 0}^{+\infty} b_n$, respectively.
If $K < +\infty$, then, from the convergence of the series $\sum_{n = 0}^{+\infty} b_n$ follows convergence of the series $\sum_{n = 0}^{+\infty} a_n$.
If $K > 0$, then, from the divergence of the series $\sum_{n = 0}^{+\infty} b_n$ follows divergence of the series $\sum_{n = 0}^{+\infty} a_n$."
The proof starts like this:
"Let $K < +\infty$ and let series $\sum_{n = 0}^{+\infty} b_n$ converge. Now, we have that for arbitrarily $\varepsilon > 0$ there exists $n_0 \in \mathbb{N}$, such that $\left| \frac{a_n}{b_n} - K \right| < \varepsilon$, for $n > n_0$ (I understand this.). From here we get the next inequality:
$$(1) \quad 0 \le a_n < (K + \varepsilon)b_n, \qquad for \quad n > n_0."$$
Unfortunately, i don't understand how did we got the inequality (1). I tried to brake starting inequality in cases and i ended up confused:
1) $\frac{a_n}{b_n} - K < 0$:
$$\left| \frac{a_n}{b_n} - K \right| < \varepsilon \iff 
-\left( \frac{a_n}{b_n} - K \right) < \varepsilon \iff$$
$$\iff \begin{cases} 
(K - \varepsilon)b_n < a_n, \quad b_n > 0, \\
(K - \varepsilon)b_n > a_n, \quad b_n < 0. 
\end{cases}$$
My question:
I didn't get the sum $K + \varepsilon$ and i currently don't have an idea how to get what is needed to continue the proof. 
Presuming that inequality (1) is correct, than the rest of the proof is understandable to me, but i don't have a full connection between two parts.
Please, could you help me with some hint or advice on how to get the inequality (1)?
Thank you, for your help and your time!
 A: (1) is just a rearrangement of the inequality above: 
$$\left|\frac{a_n}{b_n} - K\right| < \varepsilon$$ is equivalent to $$-\varepsilon < \frac{a_n}{b_n} - K < \varepsilon,$$ and (since $b_n$ is positive) multiplying everything in sight by $b_n$ gives $$-\varepsilon b_n < a_n - Kb_n < \varepsilon,$$
and adding $Kb_n$ to everything gives $$b_n(K - \varepsilon) < a_n < b_n(K + \varepsilon),$$
but we already know that $a_n$ is positive, so we have $$0 \leq a_n < b_n(K + \varepsilon).$$ 
A: Edit
$$
\left| \frac{a_n}{b_n} - K \right| < \varepsilon\iff \left| \frac{a_n}{b_n} - K \right| +K < K+\varepsilon\label{A}\tag{A}
$$
Then by the triangle inequality we have
$$
\left| \frac{a_n}{b_n} - K \right| +K \ge\left| \frac{a_n}{b_n} - K +K \right| = \left| \frac{a_n}{b_n}\right|\label{B}\tag{B}
$$
and using \eqref{A} and \eqref{B} we obtain
$$
\left| \frac{a_n}{b_n}\right|<K+\varepsilon\iff 0 \le |a_n| < (K + \varepsilon)|b_n|, \quad \text{ for }\quad n > n_0\label{C}\tag{C}
$$
Now since the coefficients of real series with positive numbers,  $a_n,b_n\ge 0$ except for a finite number of them, say
$$
\begin{split}
a_n&\ge 0\quad\text{ for all }n>n_a\\
b_n&\ge 0\quad\text{ for all }n>n_b
\end{split}\label{D}\tag{D}
$$ Inequality \eqref{A} implies that $a_n/b_n>0$ for all $n>n_0$ since $K>0$ and by a standard lemma in mathematical analysis
$$
\lim_{n\to\infty} \left| \frac{a_n}{b_n} - K \right|=0\iff\lim_{n\to\infty} \left( \frac{a_n}{b_n} - K \right)=0
$$
This implies that $a_n$ and $b_n$ have the same sign for all $n>n_0$ thus, by defining
$$
n_+=\max\{n_0, n_a, n_b\},
$$
we have that, for all $n>n_+$, the coefficients $a_n$ and $b_n$ are respectively non negative and positive real numbers so we can remove the absolute values from \eqref{C} and write
$$
0 \le a_n < (K + \varepsilon)b_n, \quad \text{ for }\quad n > n_+
$$
