# 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

• (ii) I would try to use the Cauchy criterion: for every small $r > 0$, we want an integer $N$ such that $$\left|\sum_{k = m}^{n} \epsilon_{k} x_{k}\right| < r \quad \mbox{ for } n \geq m \geq N,$$ and similarly for the $\epsilon_{n y_{n}$ series.
– avs
Commented Aug 9, 2016 at 20:50
• I am almost sure I have seen (ii) on here before or something very close to it. Commented Aug 9, 2016 at 23:18
• @mercio Indeed, (i) is almost surely a duplicate, (ii) is actually unrelated, and the pair should not have been asked in the same question.
– Did
Commented Aug 10, 2016 at 6:16

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.

• Do you know whether the same holds for countably infinitely many sequences? I don't think it does, but I don't know any counterexample. Commented Aug 10, 2016 at 11:16
• @Batominovski: Not sure about that. It doesn't hold in general Banach spaces, but of course that's not quite the same thing.
– user138530
Commented Aug 10, 2016 at 15:52

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

$\forall\epsilon>0,\exists N>0$ 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?

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 $\sum \epsilon_nx_n = 0.$