The set of all limits of sub-series of an absolute convergent series Let $\sum_{i=1}^\infty a_i$ be an absolute convergent series and let $\sum_{i=1}^\infty b_i$ be a sub-series of it, i.e.
$$b_j=c_j\cdot a_j\quad (c_j\in\{0,1\}),\qquad\forall j\in\mathbb N$$
We can say that $\sum_{i=1}^\infty b_i$ is absolutely convergent as well. So its limit, say $L$, is a real number.
Now the question is, what can be said about the set of all possible values of $L$? Is it connected? Closed maybe? Neither closed / connected / compact? I have no idea.
Can we determine whether a specific number belongs to this set or not? I was specially interested in finding a sub-series of $\sum \frac 1{n^2}$ whose limit is, say $\frac{\pi}6$, and then I ended up with this question.
 A: The set is a compact subset of $\mathbb{C}$
Let $K=\{0,1\}^{\mathbb{N}^*}$ and endow $K$ with the product topology where each $\{0,1\}$ has the discrete topology. It can be shown (even without AC/Tychonoff's Theorem) that $K$ is compact. For $i\in\mathbb{N}^*$ we denote by $p_i:K\to\{0,1\}$ the canonical projection on the $i$-th component. We recall that the topology on $K$ is the least topology that makes each $p_i$ continuous.
Let $(a_i)_{i\geq1}$ be a sequence of real (or complex) numbers such that the series $\sum_i a_i$ is absolutely convergent. Define the mapping
$$\begin{array}[t]{ccccc}\mathbf{a}&:&K&\longrightarrow&\mathbb{C}\\&&(c_i)_{i\geq1}&\longmapsto&\displaystyle\sum_{i=1}^{+\infty}a_i c_i.\end{array}$$
Since we assumed that the series $\sum_i a_i$ is absolutely convergent, it is not hard to check that the mapping $\mathbf{a}$ is well-defined (e.g., using the comparison test).
We now show that $\mathbf{a}$ is continuous: let $(c_i)_{i\geq1}\in K$ and let $\varepsilon>0$. Since the series $\sum_i a_i$ is absolutely convergent, there exists $N\in\mathbb{N}^*$ such that $\sum_{i=N+1}^{+\infty}\lvert a_i\rvert<\varepsilon$. Consider the open subset
$$O_N=\bigcap_{i=1}^N p_i^{-1}\bigl(\{c_i\}\bigr)$$
of $K$. By definition of the product topology, $O_N$ is an open subset of $K$ (indeed, $O_N$ is a finite intersection of inverse images of open sets by the continuous mappings $p_i$). Let $(d_i)_{i\geq1}\in O_N$. By definition, this means that $(d_i)_{i\geq1}$ is a sequence with values in $\{0,1\}$ and
$$\forall i\in\{1,\ldots,N\},\ d_i=c_i.$$
Then
$$\Bigl\lvert\mathbf{a}\bigl((d_i)_{i\geq1}\bigr)-\mathbf{a}\bigl((c_i)_{i\geq1}\bigr)\Bigr\rvert=\left\lvert\sum_{i=N+1}^{+\infty}(d_i-c_i)a_i\right\rvert\leq\sum_{i=N+1}^{+\infty}\lvert a_i\rvert<\varepsilon.$$
Hence $\mathbf{a}$ is continuous.
It is a standard fact that the image of a compact space by a continuous mapping is compact: we hence conclude that $\mathbf{a}(K)$ is a compact subset of $\mathbb{C}$.
A: Necessary condition for the set to be connected
Let us denote the set of all possible limits of "sub-series" by $M$. gniourf_gniourf has proven that $M$ is always a compact subset of $\mathbb{C}$. Since you also asked for connectedness:
Claim: If $M$ is connected, then the sequence $(a_i)_{i\in\mathbb{N}}$ does not become constant, or if $M = \{0\}$.
Proof: If the sequence $(a_i)$ becomes constant (i. e. $\exists\ N$ such that $a_i = 0$ for all $i > N$), then $M$ contains only finitely many values (at most $2^N$). Hence $M$ is disconnected, unless $M$ has only one element.
I feel very much like this should be an equivalence-statement, but I couldn't find a proof for the other direction so far.
