Given k unbounded sequences, must there exist a sequence such that the product of all k sequences with it is converging, diverging? Let sequences $\left( a_{n}^{(k)} \right)^{\infty}_{n=1}$, where $k \in \mathbb{N}$, so there are is a finite number of sequences, be unbounded. Must there exist a sequence $\left( X_{n} \right)^{\infty}_{1}$ such that $X_{n}>0$ and $\lim{X_{n}} = 0$ and all
1) $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ converge?
2) $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ diverge?
The easy intuitive example is $a_n^{(k)}=n^{k}$ and $X_n = \frac{1}{n^{k}}$ Then $X_n > 0$, $X_n$ converges to 0, and $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ converges (to 0 or to 1). So, what I have tried so far is this, let $ X_n = \frac{1}{\max(1, \lvert a_n^{(1)} \rvert, \lvert  a_n^{(2)} \rvert,\dots, \lvert a_n^{(k)} \rvert)^2}$. Then we have $X_n > 0$ and $\lim{X_n}=0$. But I am not sure where to go from now, intuitively, the sequence $\left(X_n a_n^{(k)}\right)^{\infty}_{n=1}$ converges to 0, because $X_n$ is always "decreasing faster", but I am not sure how to prove that this is true and that a sequence like this always exists.
The second part seems easier at first, but I am at even a greater loss there. I cannot think of an example where it would hold.
Anyone could give me some advice on how to proceed? 
 A: If there is only one sequence $(a_n)$, then $x_n=1/(n(1+|a_n|))$ is such that $x_n\gt0$, $x_n\lt1/n$ and $|x_na_n|\lt1/n$ hence $x_n\to0$ and $x_na_n\to0$. 
Likewise, let $A_n=\max\{|a_m|;m\leqslant n\}$, then by hypothesis $A_n\to\infty$ hence $x_n=1/(1+\sqrt{A_n})$ is such that $x_n\gt0$, $x_n\to0$, and, each time $(|a_n|)$ reaches a new maximum, $|a_n|=A_n$ hence, if $A_n\gt1$ then $|x_na_n|=A_n/(1+\sqrt{A_n})\gt\sqrt{A_n}/2$, in particular, $(x_na_n)$ is unbounded.
If there are $k$ sequences $(a^{(i)}_n)$ for $1\leqslant i\leqslant k$, assume that, for each $i$, $(x^{(i)}_n)$ is such that $x_n^{(i)}\gt0$, $x_n^{(i)}\to0$ and $x_n^{(i)}a_n^{(i)}\to0$. Let $x_n=\min\{x_n^{(i)};1\leqslant i\leqslant k\}$, then $x_n\gt0$, $x_n\to0$ and $x_na_n^{(i)}\to0$ for each $1\leqslant i\leqslant k$.
Likewise, assume that, for each $i$, $(x^{(i)}_n)$ is such that $x_n^{(i)}\gt0$, $x_n^{(i)}\to0$ and $(x_n^{(i)}a_n^{(i)})$ is unbounded. Let $x_n=\max\{x_n^{(i)};1\leqslant i\leqslant k\}$, then $x_n\gt0$, $x_n\to0$ and $(x_na_n^{(i)})$ is unbounded, for each $1\leqslant i\leqslant k$.
