Let $a_n$, $b_n$ be bounded sequences of real numbers and suppose that there exists N so that $a_n$ $\leq$ $b_n$ for all n $\geq$ N. Prove that lim sup $a_n$ $\leq$ lim sup $b_n$.
I can see why this would be true.
We are given that $a_n$ $\leq$ $b_n$. Since sup $a_n$ is just another term in our sequence $a_n$ and sup $b_n$ is just another $b_n$, then sup $a_n$ $\leq$ sup $b_n$.
I also understand that sup $a_n$ is not only an upper bound, but the least upper bound of ($a_n$). So $a_n$ $\leq$ sup $a_n$. Similarly, $b_n$ $\leq$ sup $b_n$.
Altogether, we have $a_n$ $\leq$ sup $a_n$ $\leq$ $b_n$ $\leq$ sup $b_n$.
Clearly, sup $a_n$ $\leq$ sup $b_n$.
But I don't know how to show that lim sup $a_n$ $\leq$ lim sup $b_n$.
I know lim sup $a_n$ is the same thing as sup $a_n$ converging to a point, let's call it $a$. So, lim sup $a_n$ means if we let $\epsilon$ > 0, then there exists N such that |sup $a_n$ – $a$| < $\epsilon$ whenever $n$ >N.
Another definition of the lim sup that may be more helpful is: Let E be the set of all subsequential limits of a bounded sequence ($a_n$), then lim sup $a_n$ = sup E. So the lim sup is the least upper bound of all the limits that the subsequences converge to.
We also have that our sequences ($a_n$) and (bn) are bounded which means that there exists M>0 such that |$a_n$| $\leq$ M for all n in N. Similaryly, there exists L>0 such that |$b_n$| $\leq$ L for all $n$ in N.
Since we're given that these sequences are bounded we should be using that in our proof.
So, I basically have all these tools – bounded sequences, convergence, etc- but I'm not sure how that all proves that lim sup $a_n$ $\leq$ lim sup $b_n$.
Best Answer
You just go to the definition of $\limsup$. We define $A_n = \sup_{m\geq n} a_m$ and $B_n = \sup_{m\geq n}b_m$. Since $a_n \leq b_n$, $A_n \leq B_n$ for every $n>N$. Then, $\inf A_n \leq A_n \leq \inf B_n \leq B_n$, but $\inf A_n = \limsup a_n$ and $\inf B_n = \limsup b_n$.