Let $a_{i}$ be a sequence of real numbers and suppose that $\limsup a_{i}$ is finite. Let $c_{i}$ be another sequence and suppose $c_{i}$ converges to $c$. Prove that if $c \ge 0$, then $\limsup c_{i} a_{i} = c \limsup a_{i}$ .
I worked out the case where $c_{i}$ is non-negative. Not sure what to do if both $a_{i}$ and $c_{i}$ are negative. Thanks!
In fact if we allow both $c_i$ and $a_i$ to be negative and have no restrictions other than the ones given, the statement is not true.
Take $c_i = -\frac{1}{i}$ (so that $c_i\rightarrow 0$) and $a_i = 0$ if $i$ is odd while $a_i = -i$ if $i$ is even. Then $\limsup a_i = 0$, but $\limsup c_ia_i = 1$.
If we require also that $\liminf a_i$ is finite, then $c_i\rightarrow 0$ implies that $a_ic_i\rightarrow 0$. If $c_i\rightarrow c>0$ then $c_i$ is eventually positive, which is the case you already know.
