$x_n$ doesn't converge to zero and is always smaller than zero and a properly divergent sequence $y_n$ There's a sequence $y_n\rightarrow\infty$ for $n\rightarrow\infty$.
Proof that if $x_n$ doesn't converge to zero and for all $n\in\mathbb{N}$ : $x_n<0$
then there's a subsequence $x_{n_j}$ of $x_n$ such that $x_{n_j}y_{n_j}\rightarrow-\infty$ for $j\rightarrow\infty$.
I have no clue how to proof this. The subsequence thing confuses me so much. If all values of $x_n$ are smaller than zero and are multiplied with $y_n$ shouldn't it automatically converge to $-\infty$?
Thanks for your help in advance!!
 A: Since $(x_n)_n$ doesn't converge to zero, there exists $\varepsilon > 0$ and a subsequence $(x_{n_j})_j$ such that $|x_{n_j}| \ge \varepsilon$ for all $j \in \Bbb{N}$.
Since $x_{n_j} < 0$, it has to be $x_{n_j} \le -\varepsilon$. For large enough $j$ we have $y_{n_j} \ge 0$ so $$x_{n_j}y_{n_j} \le -\varepsilon y_{n_j} \xrightarrow{j\to\infty} -\infty.$$
A: Consider $y_n=n, x_n=-\frac1n$. We easily see that $x_ny_n$ doesn't automatically diverge to $-\infty$ just because $x_n$ are negative. This is the reason we can't have $x_n\to0$.
Next, consider $y_n=n$ and
$$
x_n=\cases{\frac1n& if $n$ is even\\-1& otherwise}
$$
Take a look at the sequence $x_ny_n$ to see why we need to work with a subsequence.
As for the proof itself, consider this: the fact that $x_n$ does not converge to $0$ (in conjunction with $x_n<0$) means

There is an $\varepsilon < 0$ such that there are infinitely many terms that fulfill $x_n<\varepsilon$

Can you guess, just from reading the above paragraph (and the fact that this is the hint I give) what a suitable subsequence might be?
