$\sum_\limits{n=1}^\infty {a_n}$ 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.
 A: Personally, I would try to prove your equivalence by thinking about the definition of infinite series. That is, by definition, $$\sum_{n=1}^\infty = \lim_{N\to\infty} S_N$$
where $S_N$ is the partial sum of the first $N$ elements in the sequence,
$$S_N=\sum_{n=1}^Na_n$$
Then, the first sum is the limit of the sequence $S_1,S_2,S_3\dots$, and, because of the definition of the subsequence $n_k$, the second sum is the limit of a subsequence of $S_1,S_2,\dots$ with all repetitions removed. (note: this statement needs a formal proof that is not hard).
Now, you only need to prove that removing all repetitions from a sequence does not change it's convergence, a much simpler task. If $b_n$ is a sequence, and $b_{n_k}$ is a sub-sequence of $b_n$ without repetitions, then:


*

*Say $b_n$ converges. Then $b_{n_k}$ is a subsequence of $b_n$ and it converges.

*Say $b_{n_k}$ converges to $b$ and let $\epsilon > 0$. Then there exists a value $K$ such that if $k>K$, $|b_{n_k}-b|<\epsilon$. Set $N = n_K$. Then, for all $n>N$, you have $b_n = b_{n_k}$ for some $k>K$ (why?), and thus $|b_n-b| < \epsilon$, and convergence is shown.

A: 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.
