# Real Analysis Proof Continuity - lim sup anbn≤(lim sup an)(lim sup bn) and …

1) If an ≥ 0 and bn ≥ 0, prove that lim sup anbn ≤ (lim sup an)(lim sup bn)

2) If {an} and{bn} are non-negative sequences and {bn} converges, prove that lim sup anbn = (lim sup an)(lim bn).

I am not sure how to show these two similar proofs. Please show details so I can understand the process.

Thank you.

• Do you have a good understanding of what the $\limsup$ is? Please, write down your definition of $\limsup$ in the question, since there are different definitions of it. – Crostul Feb 17 '15 at 19:52

Let be $n\geq0$. For all $k\geq n$ , one has$$a_{k}\leq\sup_{k\geq n}a_{k}$$ and also$$b_{k}\leq\sup_{k\geq n}b_{k}.$$ Since everything is non-negative, one gets$$a_{k}b_{k}\leq\left(\sup_{k\geq n}a_{k}\right)\left(\sup_{k\geq n}b_{k}\right)$$ whence$$\sup_{k\geq n}a_{k}b_{k}\leq\left(\sup_{k\geq n}a_{k}\right)\left(\sup_{k\geq n}b_{k}\right).$$ To conclude, one takes the limit $n\rightarrow+\infty$.