May I use the triangle inequality for infinite series? I have to prove the following statement:

Let $\lim_{n\to \infty}r_n=0$. Show that $\forall\varepsilon>0 \ \ \ \exists \, n_0 \in \mathbb N \ \ \ \forall x \in(-1,1):$
  $$\left\lvert \sum\limits_{n=n_0}^\infty r_nx^n \right\rvert< \frac{\varepsilon}{1-|x|}$$
  Conclude that the series converges absolutely for all $|x|<1$.

Here is what I did:
Let $\varepsilon>0$. Since $\lim_{n\to \infty}r_n=0$, there exists such $n_0 \in \mathbb N$ that $|r_n|<\varepsilon$.
\begin{align} 
\left\lvert \sum\limits_{n=n_0}^\infty r_nx^n \right\rvert &\leq \sum\limits_{n=n_0}^\infty |r_n| |x^n|
 \\ &\leq \sum\limits_{n=n_0}^\infty \varepsilon |x^n|
 \\ &=  \varepsilon \frac{|x|^{n_0}}{1-|x|}
 \\ &<  \frac{\varepsilon}{1-|x|}
\end{align}
My question is about the first line: is this correct? I know that it would be correct if we knew that the series converges absolutely but at that point we can't assume the absolute convergence since this is exactly what needs to be concluded later on.
Any help is greatly appreciated.
 A: Yes, this is correct. Recall that an infinite series is actually a limit. So, the first line is actually $$\left|\lim_n\sum_{k\le n}a_k\right|=\lim_n\left|\sum_{k\le n}a_k\right|$$ for some $a_k$. Now, for each $n$, $$ \left|\sum_{k\le n}a_k\right|\le\sum_{k\le n}|a_k|,$$ where the last inequality is the triangle inequality.
But $$\sum_{k\le n}|a_k|\le\lim_{m\to\infty}\sum_{k\le m}|a_k|=\sum_k |a_k|.$$ Since each term in the sequence of partial sums $|\sum_{k\le n}a_k|$ is at most $\sum_k |a_k|$, it follows that their limit $|\sum_k a_k|$ is bounded by the same quantity as well. Notice that this only assumes convergence (rather than absolute convergence) of the sequence, as the argument works (trivially) if $\sum_k|a_k|=+\infty$.
A: You can argue that the series converges absolutely pretty simply. Since $\lim_{n\rightarrow\infty}r_n = 0$, there exists $n_1$ such that for all $n > n_1$, $|r_n| < 1$. Then you can argue that $\sum_{n=n_1}^{\infty} |r_n x^n| < \sum_{n=n_1}^{\infty} |x|^n$. Thus, that part of the series converges and since there are finitely many $n$ between $n_0$ and $n_1$, the overall series converges. Everything else is right (though you should change the $\leq$ to $<$ in the second line of your proof).
