If the series is absolutely convergent, then it is also conditionally convergent. Let $\sum_{n=m}^{\infty}a_{n}$ be a formal series of real numbers. If the series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality:
\begin{align*}
\left|\sum_{n=m}^{\infty}a_{n}\right| \leq \sum_{n=m}^{\infty}|a_{n}|
\end{align*}
MY ATTEMPT (EDIT)
Since $S_{n} = |a_{m}| + |a_{m+1}| + \ldots + |a_{n}|$ converges, it is a Cauchy sequence.
Therefore for every $\varepsilon > 0$, there exists a natural number $N\geq m$, such that
\begin{align*}
p \geq q\geq N \Longrightarrow |a_{p} + a_{p-1} + \ldots + a_{p-q+1}| \leq |S_{p} - S_{q}| \leq \varepsilon
\end{align*}
whence we conclude that $s_{n} = a_{m} + a_{m+1} + \ldots + a_{n}$ converges, because it is Cauchy too.
Similar reasoning proves that $|a_{m} + a_{m+1} + \ldots + a_{N}|$ converges. 
This is because $||x|-|y|| \leq |x - y|$.
Consequently, one has that
\begin{align*}
\left|\sum_{n=m}^{N}a_{n}\right| \leq \sum_{n=m}^{N}|a_{n}| \Longrightarrow \lim_{N\rightarrow\infty}\left|\sum_{n=m}^{N}a_{n}\right| \leq \lim_{N\rightarrow\infty} \sum_{n=m}^{N}|a_{n}| \Longrightarrow \left|\sum_{n=m}^{\infty}a_{n}\right| \leq \sum_{n=m}^{\infty}|a_{n}| 
\end{align*}
and we are done.
Could someone please double-check my arguments?

Most definitions I know of conditional convergence states (e.g. https://en.wikipedia.org/wiki/Conditional_convergence) 

A series is conditionally convergent if it is convergent but not absolutely convergent. 

Which means also that if a series is absolutely convergent, it cannot be the case that it is also conditionally convergent. 
