Science:Math Exam Resources/Courses/MATH220/April 2011/Question 10 (b)/Solution 1

The sequence {a_n} is bounded, which means that there exists a real number M such that

$\displaystyle |a_{n}|\leq M\quad {\text{for all }}n\geq 1$

Hence the supremum of all the numbers a_n is at most M and by definition of the numbers b_n we have that

$\displaystyle |b_{n}|\leq M\quad {\text{for all }}n\geq 1$

or in other words, the sequence {b_n} is bounded as well. That sequence will be converging if it is decreasing, that is if

$\displaystyle b_{n+1}\leq b_{n}\quad {\text{for all }}n\geq 1$

Which we can easily show to be true. Indeed, if we denote by

$\displaystyle B_{n}=\{a_{m}\mid m\in \mathbb {N} {\text{ s.t. }}m\geq n\}\quad {\text{for all }}n\geq 1$

then

$\displaystyle b_{n}=\sup(B_{n})\quad {\text{for all }}n\geq 1$

and clearly

$\displaystyle B_{n+1}\subseteq B_{n}\quad {\text{for all }}n\geq 1$

and by part (a) we have that

$\displaystyle b_{n+1}=\sup(B_{n+1})\leq \sup(B_{n})=b_{n}$

This concludes our proof.

Advanced note: We call the limit of this convergent sequence the limit superior of the original sequence {a_n}.