Proof of $\lim\sup(a_nb_n)\leq \lim\sup(a_n)\limsup(b_n)$

Let $a_n>0$ and $b_n\geq 0$, then $\lim\sup(a_nb_n)\leq \lim\sup(a_n)\limsup(b_n)$

My attempt at a proof is as follows. Let $A_n=\sup\{a_n, a_{n+1},…\}$, $B_n=\sup\{b_n, b_{n+1},…\}$, and $C_n=\sup\{a_nb_n, a_{n+1}b_{n+1},…\}$.

Note: $a_mb_m \leq A_nB_n$ for all $m \geq n$.

Thus $\limsup(a_nb_n)=\lim C_n \leq \lim (A_nB_n) = (\lim A_n)(\lim B_n) = (\limsup a_n)(\limsup b_n).$

Solutions Collecting From Web of "Proof of $\lim\sup(a_nb_n)\leq \lim\sup(a_n)\limsup(b_n)$"

Missing pieces:

  1. You need to assume the sequences are bounded. Otherwise the statement may be even nonsensical: how to interpret $1\le 0\cdot \infty$, for example?

  2. $(A_n)$ and $(B_n)$ are nonnegative nonincreasing sequences; these properties imply the existence of the limits $\lim A_n$ and $\lim B_n$, which allows to conclude about the existence and the value of $\lim (A_nB_n)$.

You could have assumed $a_n\ge 0$ instead of $a_n>0$.