The sequence {an} is bounded, which means that there exists a real number M such that
![{\displaystyle \displaystyle |a_{n}|\leq M\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/679228030aa2bb4672772ed66f32cc944ec31afd)
Hence the supremum of all the numbers an is at most M and by definition of the numbers bn we have that
![{\displaystyle \displaystyle |b_{n}|\leq M\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f51f1804de61a9f66d485c9d093a14f840378c59)
or in other words, the sequence {bn} is bounded as well. That sequence will be converging if it is decreasing, that is if
![{\displaystyle \displaystyle b_{n+1}\leq b_{n}\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/56f247fd0379ca48c8556a2996da1ff35f498180)
Which we can easily show to be true. Indeed, if we denote by
![{\displaystyle \displaystyle B_{n}=\{a_{m}\mid m\in \mathbb {N} {\text{ s.t. }}m\geq n\}\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2b6ad6f8d2cb8232703c24cc3903e88091f72b37)
then
![{\displaystyle \displaystyle b_{n}=\sup(B_{n})\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e84507426d1fc7fe4b52ff65f7181c8206850285)
and clearly
![{\displaystyle \displaystyle B_{n+1}\subseteq B_{n}\quad {\text{for all }}n\geq 1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/aff313a23114fcd40c3646b3809f6852975fbdd8)
and by part (a) we have that
![{\displaystyle \displaystyle b_{n+1}=\sup(B_{n+1})\leq \sup(B_{n})=b_{n}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1147361dc3818dae8d30073fc1121c5944092037)
This concludes our proof.
Advanced note: We call the limit of this convergent sequence the limit superior of the original sequence {an}.