Getting close to any real with a semi-convergent series I got a really intuitive exercise here, but I'm having a hard time actually fulfilling a proof.
Let $\sum_{0}^{}a_n$ be a semi-convergent real series.
Let $l \in \mathbb R$.
Show that there exists a sequence $e_n \in \left\{ -1,1 \right\} ^{ \mathbb\ N } $ such that $$\sum_{0}^{\infty}e_na_n=l$$


*

*One important point is the semi-convergence : it implies that $\sum_{0}|a_n| \to +\infty$

*Since $\sum_{0}^{}a_n$ is convergent, $a_n \to 0$
What I'd like to do is to sum some positive $|a_i|$ to go past $l$ and then sum some negative $(-1)|a_i|$ to back before $l$. Since $a_n$ converges to $0$, one may come very close to $l$ repeating the process.
That is not a proof whatsoever though.
Thanks for your help. 
 A: Let $(b_n)_{n=0}^\infty$ be a sequence of nonnegative numbers such that $b_n\to 0$ and $\sum_{n=0}^\infty b_n=+\infty$. Let $l\in\mathbb R$.
Define a sequence $(e_n)_{n=0}^\infty$ with $e_n\in\{-1,1\}$ recursively as follows:
For $n\in\mathbb N_0$, assume that all $e_k$ with $k<n$ are already defined.
Let $s_n=\sum_{k=0}^{n-1} e_k b_k$ and define
$$ e_n=\begin{cases}+1&\text{if }s_n<L,\\-1&\text{if }s_n\ge L.\end{cases}$$
Then we have $e_n=+1$ for infinitely many $n$ because otherwise we'd have $e_n=-1$ for all $n>N$ for some $N$ and hence $L\le s_n=s_N-\sum_{k=N}^{n-1}b_k$, then $\sum_{k=0}^{n-1}b_n\le \sum_{k=0}^{N-1}b_k+s_N-L$ and finally $\sum_{n=0}^\infty b_n\le \sum_{k=0}^{N-1}b_k+s_N-L$, contradiction.
By a similar argument, we have $e_n=-1$ for infinitely many $n$.
Now to formally show that $\sum_{n=0}^\infty e_nb_n=\lim_{n\to\infty} s_n$ is indeed $=L$ proceed as follows:
For $\epsilon>0$ first pick $N_0$ such that $b_n<\epsilon$ for all $n>N_0$. Then pick $N>N_0$ such that $e_N\ne e_{N+1}$ (which is possible beacuse both signs occur infinitely often). Now show by induction that $|s_n-L|<\epsilon$ for all $n>N$.
