If $\lim_{n\to \infty}a_n = a\in \mathbb{R}$ . Prove that $\limsup_{n\to \infty}a_n x_n=a\limsup_{n\to \infty}x_n$ .

Question

Note: $x_n$ is a sequence which is not necessarily convergent.

The following was my attempt.
Since $\lim_{n\to \infty}a_n=a$ then $\limsup_{n\to \infty}a_n=a$ .
Also $\sup(a_nx_n)=\sup(a_n)\sup(x_n)$.
Therefore, $\limsup_{n\to \infty}a_nx_n=\limsup_{n\to \infty}a_n\limsup_{n\to \infty}x_n=a\limsup_{n\to \infty}x_n$.

I am not sure if this is correct. So could someone please show me how it is done. Thanks

Answer 1

For this $\lim\sup$ should not be $\infty$. Also I will assume $\lim$ and $\lim\sup$ are positive. I will do the lower bound and you should carry out the rest of the argument for the upper bound. By definition of limit and $\lim\sup$ given $\epsilon>0$ there exists $N$ such that for all $n>N$ $a_n>(1-\epsilon)a$ and for all $N$ there exists $m>N$ such that $x_m>(1-\epsilon)\lim \sup x_n$. Combining these you find $$a_mx_m>(1-\epsilon)^2 a\lim \sup x_n$$

We can always find arbitrarily large $m$ satisfying this hence $\lim\sup a_nx_n\geq (1-\epsilon)^2a \lim \sup x_n$. Now let $\epsilon\rightarrow 0$.

Answer 2

How about $a_n = \frac{1}{n}$ and $x_n = n$?

Then $\sup a_nx_n = 1$ do we have $0 \times +\infty =1$, but then, what about $b_n = \frac{2}{n}$

Answer 3

Your claim that $\sup a_n x_n = \sup a_n \sup x_n$ isn't true, even for positive convergent sequences. Take (for instance) $a_n = \frac{n+1}{n}$ and $x_n = \frac{n}{n+1}$.

The proof can't be fixed because as stated the result isn't true unless the terms are nonnegative. Take $a_n = -1$ and $x_n = (-1)^n$.