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
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.
$\lim_\limits{n\to \infty} x_n = 0$

$\forall\epsilon&gt;0,\exists N&gt;0$ such that $n&gt;N\implies |x_n|&lt;\epsilon.$

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 $N&gt;n,$ we can choose the sign of $e_{n+1}$ such that if $M_n&gt;M_N, e_{n+1}x_{n+1} &lt; 0$ and if $M_n &lt; M_N, e_{n+1}x_{n+1} &gt; 0.$ 
With $e_{n+1}$ chosen in this way, since $|x_{n+1}| &lt; \epsilon, |M_{n+1} - M_N| &lt; \epsilon$

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?