Change signs of sequence elements such that sum of elements converges Suppose there is a sequence like $(a_i)_{i \ge 1}$, such that infinite sum of sequence diverges and for every element $|a_i|\le1$ and $a_n \to 0$. Is it possible for any sequence to change sign of each element in a way that their sum converges? also is it possible to make it converge to zero or make it bounded?
I thought about Harmonic series and the answer for it ,is positive (to converge I mean), however I am not sure if this is correct in general, I really appreciate if you could help me.
 A: If $(a_n)$ is a sequence of real numbers such that $a_n \to 0$ and  $\sum_{n=1}^\infty a_n$  diverges then there is a sequence of “signs” $(s_n)$ such that $s_n = \pm 1$ for all $n$ and $\sum_{n=1}^\infty  s_n a_n = 0$.
Here is a possible construction per induction. Without loss of generality we can assume that all $a_n \ge 0$. We start by setting $s_1 = +1$. If $s_1, \ldots,s_n$ are already defined then we set
$$
 s_{n+1} = \begin{cases}
 -1 & \text{ if } \sum_{k=1}^n  s_k a_k \ge 0 \\
 +1 & \text{ if } \sum_{k=1}^n  s_k a_k < 0 \, .
\end{cases}
$$
Now we show that $\sum_{n=1}^\infty  s_n a_n = 0$. Let $\epsilon > 0$. Since $a_n \to 0$ there is an index $N$ such that $0 \le a_n < \epsilon $ for $n \ge N$. Set $S = \sum_{k=1}^N  s_k a_k$. The following partial sums decrease (if $S \ge 0$) or increase (if $S < 0$) until they reach a value in the range $[-\epsilon, \epsilon]$ and then stay in that range. So there is a $M > N$ such that $-\epsilon \le \sum_{k=1}^n  s_k a_k \le \epsilon$ for all $n \ge M$. Such an index $M$ exists for all $\epsilon > 0$, and that proves the converges of $\sum_{k=1}^n  s_k a_k$ to zero.
In the same way one can show that for every $L \in \Bbb R$ there is a choice of signs $(s_n)$ such that $\sum_{n=1}^\infty  s_n a_n = L$.
