real analysis - $sum_limits{n=1}^infty {a_n}$ converges $iff sum_limits{n=1}^infty {a_{n_k}}$ converges.

Let $({a_n})_{n\in{\mathbb{N}}}$ a sequence,and let $({a_{n_k}})_{k\in{\mathbb{N}}}$ the sequence of all terms of $({a_n})$ different than zero. Then $$\sum_\limits{n=1}^\infty {a_n}\text{ converges} \iff \sum_\limits{n=1}^\infty {a_{n_k}} \text{ converges}$$

My approach to the proof:

$\Rightarrow$ Suppose $\sum_\limits{n=1}^\infty {a_n}$ converges, so for every $\epsilon>0$ ,$\exists N\in{\mathbb{N}} $ so that $\forall n,m\ge N$ then $|{a_m}-{a_n}|<\epsilon$.
Let $B=[j\in{\mathbb{N}}|{a_j}=0]$, so that ${a_n}={a_j}\cup {a_i}$. Can I express $\sum_\limits{n=1}^\infty {a_n}$ as $\sum_\limits{n\notin B}^\infty {a_n} +\sum_\limits{n\in B}^\infty {a_n}$

I need some help proving this, it might be trivial but I'm having problems with notation. Any help will be appreciated.

Answer

Define the partial sums

$$
S_n = \sum_{i=1}^n a_i
$$

For any given $\varepsilon$, let $N$ be that integer such that for all $p, q \geq N, |S_p-S_q| < \varepsilon$. Either there exists a minimum $K$ such that $n_K \geq N$ (and then for all $r, s \geq K, |S_{n_r}-S_{n_s}| < \varepsilon$ and the second series is convergent), or else there does not exist such a minimum $K$, in which case the second series has a finite number of terms and is convergent.

ETA: Oh yes, the inverse. For any given $\varepsilon$, let $K$ be that integer such that for all $r, s \geq K, |S_{n_r}-S_{n_s}| < \varepsilon$. We observe that for any $n$, $S_n = S_{n_r}$ where $r = \max_{n_s \leq n} s$ (since we are only adding a finite number of trailing zeros). Let $N = n_K$. Then for any $p, q \geq N, |S_p-S_q| = |S_{n_r}-S_{n_s}| < \varepsilon$ for some $r, s \geq K$, and the first series is convergent.