Must there be a sequence $(\epsilon_n)$ of signs such that $\sum\epsilon_nx_n$ and $\sum\epsilon_ny_n$ are both convergent? 
Let $(x_n)$ and $(y_n)$ be real sequences.
(i) Suppose $x_n \rightarrow 0$ as $n \rightarrow \infty.$ Show that
  there is a sequence $(\epsilon_n)$ of signs (i.e., $\epsilon_n \in
 \{−1, +1\}$ for all $n$) such that $\sum \epsilon_nx_n$ is convergent.
(ii) Suppose $x_n \rightarrow 0$ and $y_n \rightarrow 0.$ Must there
  be a sequence $(\epsilon_n)$ of signs such that $\sum\epsilon_nx_n$
  and $\sum\epsilon_ny_n$ are both convergent?

I'm struggling to come up with formal proofs, for (i) I've seen that we simply pick a limit and and then as soon as our sum passes the limit we set $\epsilon_n=-1$ until we pass it again and so on, oscillating about the limit but as $x_n \rightarrow 0$ we converge to it. for (ii) I don't think there must be such a sequence of $\epsilon_n$ but I can't construct a proof or counter example. So I would ask for a solution to (ii) and possibly a better way of constructing answers/tackling these problems in general.
Thank you
 A: The answer to (ii) is yes, there is such a sequence of signs $\epsilon_n$. See Theorem 2.2.1 here, where this result (for any number of series, formulated for vector valued series) is referred to as the Dvoretzky-Hanani Theorem.
A: $\lim_\limits{n\to \infty} x_n = 0$
$\forall\epsilon>0,\exists N>0$ such that $n>N\implies |x_n|<\epsilon.$
I don't what to make poor epsilon pull double duty.  I am going to call the sequence of signs e.
$M_n = \sum_\limits {i=1}^n e_i x_i$
For the first N elements we can let $e_i$ be strictly positive.
When $N>n,$ we can choose the sign of $e_{n+1}$ such that if $M_n>M_N, e_{n+1}x_{n+1} < 0$ and if $M_n < M_N, e_{n+1}x_{n+1} > 0.$ 
With $e_{n+1}$ chosen in this way, since $|x_{n+1}| < \epsilon, |M_{n+1} - M_N| < \epsilon$
What about ii) my intuition says that it can't be done.
Suppose $\sum |y_n|$ and $\sum |x_n|$ diverge, but the sequences are going to zero.
for any sequence of signs $e_n$ such that $\sum e_n x_n$ converges, there exists a sequences of signs $f_n$ such that $e_n f_n y_n = |y_n|$  and $f_ny_n$ would still be a sequence that is going to zero.
For all convergent $\sum e_n x_n$ there exists a $y_n$ such that $y_n\to 0$ and $\sum e_n y_n$ is divergent. 
But that is not quite the same thing as we have been asked to prove, is it?
A: Sketch (omitting some details): If $\sum |x_n| < \infty,$ we're done. So assume $\sum |x_n| = \infty.$ Let $n_1 = 1.$ Then let $n_2$ be the smallest $n$ such that $|x_1|-( |x_2| +\cdots + |x_{n_2}|) < 0.$ We know there will be such an $n_2$ because $\sum |x_n| = \infty.$ Then let $n_3$ be the smallest $n$ such that
$$|x_1|-( |x_2| +\cdots + |x_{n_2}|) + (|x_{n_2+1}|+ \cdots |x_{n_3}|) >0.$$
Keep going. Because $x_n \to 0,$ we have
$$\sum_{k=1}^{\infty}(-1)^{k+1} \sum_{n=n_k}^{n_{k+1} - 1}|x_n| =0.$$
This will yield a sequence $\epsilon_n$ such that $\sum \epsilon_nx_n = 0.$
A: (Just a comment on (i))
Your answer to (i) is incorrect how you stated it. It could be that $x_n \to 0$ but they go too quickly to approach any limit (this happens if $\sum_n |x_n| < \infty$). For example, take $x_n = \frac{1}{2^n}$, and say you pick the limit $L = 3$. Then, according to what you said, you keep taking $\epsilon_i = 1$ positive terms until your sum passes $L$...except that you never get past $L$. The sum never reaches $2$. You can make this correct if you handle the case that $\sum_{i} |x_i|$ converges separately. Just make sure you write out all the details of the proof; as it is you seem to be asserting you can make the limit $L$ for any real $L$, which is not necessarily true. Also make sure not to assume that $x_i$ is positive.
