Does limsup(x_n·y_n) = (lim x_n)·(liminf y_n) when lim x_n < 0?

[zzzhhh]

If $$x_n\geq 0, y_n\geq 0$$ and $$\lim \limits_{n \to \infty }x_n$$ exists, we have $$\limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\limits_{n\to\infty}y_n)$$. But if $$\lim\limits_{n\to\infty}x_n<0$$, do we have analog equation(I guess $$\limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\liminf\limits_{n\to\infty }y_n)$$)? and what change should be made to conditions to achieve the analog equation? Formal source of reference such as textbooks or webpages is recommended. Thanks!

 

[Gib Z]

Do you know how to prove the first result? Make a proof, take a careful look to see where you used the assumption that x_n is positive, and then it should be clear how things become changed when you assume x_n is negative.

 

[Landau]

As always with liminf and limsup, use the sign trick!
$$\liminf (-a_j)=-\limsup(a_j).$$
If x_n<0, then -x_n>0, so we can apply your result (in the form $\liminf(a_jb_j)=\lim(a_j)\liminf(b_j)$ for a_j,b_j nonnegative and a_j convregent, i.e. sup replaced by inf) to get:

$$\limsup (x_ny_n)=-\liminf (-x_ny_n)=-\lim(-x_n)\liminf(y_n)=\lim(x_n)\liminf(y_n)$$

 

[zzzhhh]

Thank you Landau, it's really a good idea!