# $\sum_{n=1}^\infty {a_n}$ is absolutely convergent and ${b_n}$ is any subsequence of ${a_n}$, then $\sum_{n=1}^\infty {b_n}$ is abs. convergent

If $\sum_{n=1}^\infty {a_n}$ is absolutely convergent and ${b_n}$ is any subsequence of ${a_n}$, then $\sum_{n=1}^\infty {b_n}$ is absolutely convergent.

My attempt of proof: Let ${b_j}={a_{n_j}}$ Then: $$|{b_{j+1}}|+|{b_{j+2}}|+...+|{b_k}|=|{a_{n_{j+1}}}|+|{a_{n_{j+2}}}|+...+|{a_{n_k}}|$$ Because $\sum_{n=1}^\infty {a_n}$ is absolutely convergent, then $\lim_{n\rightarrow\infty} |{a_{n+1}}|+|{a_{n+2}}|+...+|{a_m}|=0$

I know that $\lim_{j\rightarrow\infty} |{b_{j+1}}|+|{b_{j+2}}|+...+|{b_k}|$ should be zero too , but I'm having trouble with notation , how can I finish my proof? Any help will be appreciated.

If $b_n$ is a subsequence of $a_n$, then there is some subsequence of $\mathbb{N}$ such that $b_k = a_{n_k}$. Let $I= \{n_k\}$.
Define the sequence $a_n' = \begin{cases} a_n, & n \in I \\ 0, & \text{otherwise}\end{cases}$. We have $|a_n'| \le |a_n|$, of course.
We see that $\sum_{k=1}^N b_k = \sum_{k=1}^N a_{n_k} = \sum_{i=1}^{n_N}a_i'$ and $\sum_{k=1}^N |b_k| = \sum_{k=1}^N |a_{n_k}| = \sum_{i=1}^{n_N}|a_i'| \le \sum_{i=1}^{n_N}|a_i| \le \sum_{i=1}^{\infty}|a_i|< \infty$.
Hence the $b_k$ are absolutely convergent.