Upper limits inequality I am trying to show that for two sequences $f_n,g_n$ that are non-negative and bounded that $ \limsup(f_ng_n) \le (\limsup (f_n))(\limsup (g_n)) $. I have come up with the example where $f_n$ is a repeating sequence $\{1,2,3,4\}$ and $g_n$ is a repeating sequence $\{(1/2),(1/3),(1/4),(1/5)\}$ then  $ \limsup(f_ng_n)= 4/5 $ but $(\limsup (f_n))(\limsup (g_n)) = 2 $. Now to put this in words it seems like the product of largest sub-sequential limit of each sequence is larger or equal to the largest sub-sequential limit of the multiples of the $n$'th entries for  $ \limsup(f_ng_n)$ but I am unsure how to express this algebraically. 
Essentially that the largest sub-sequential limits of each sequence may be larger than the largest sub-sequential of the product $f_ng_n$.
 A: For each $k\ge n\in\mathbb N$ 
$$f_kg_k\leq \sup\limits_{k\ge n}{f_k}\sup\limits_{k\ge n}{g_k}$$
Now the right hand side is a fixed number and you can take supremum on both sides over $k\ge n$ and so you get
$$\sup\limits_{k\ge n}{f_kg_k}\leq \sup\limits_{k\ge n}{f_k}\sup\limits_{k\ge n}{g_k}$$
Finally you let $n\to\infty$ and the result follows by the definition of $\limsup$.
Note that as both sequences are bounded from above and below by $0$ so is the sequence of the products $f_kg_k$ and therefore $\sup\limits_{k\ge n}{f_kg_k}$ exists for each $n$ and the sequence $\sup\limits_{k\ge n}{f_kg_k}$ is decreasing $\Rightarrow$ the limit $\lim\limits_{n\to\infty}{\sup\limits_{k\ge n}{f_kg_k}}$ exists. The same goes for the limit on the right: the sequence $\sup\limits_{k\ge n}{f_k}\sup\limits_{k\ge n}{g_k}$ is bounded from above, and from below by $0$, and decreasing $\Rightarrow$ has a limit. Also, because each of the sequences $\sup\limits_{k\ge n}{f_k}$ and $\sup\limits_{k\ge n}{g_k}$ possesses a limit, the limit of the product is equal to the product of the limits.
A: Let $A_n=\sup \{ f_n,f_{n+1},\dots \}$, $B_n=\sup \{ g_n,g_{n+1},\dots \}$ and $C_n=\sup\{ f_n g_n,\dots\}$
Since $f_k g_k\leq A_n B_n$ for any $k\geq n$ we have 
$$
\lim\sup (f_n g_n)=\lim C_n\leq \lim(A_nB_n)=\lim A_n\lim B_n=\lim\sup f_n\lim\sup g_n
$$
