Let $(a\_{n})\_{n}$ be a bounded real-valued sequence. Suppose that $(b\_{n})\_{n}$ is a sequence (not necessarily a subsequence) in the set $A:=(a\_{n})\_{n}$. Assume that the limit $\lim\limits\_{n}b\_{n}$ exists. Moreover, assume that $(b\_{n})\_{n}$ is not eventually constant (we say that a sequence $(c\_{n})\_{n}$ is eventually constant if there exists a positive integer $N$ so that $c\_{n}=c\_{N+1}$ for all $n>N$). My question is the following:


Question. $\lim\limits\_{n}b\_{n}\leq \limsup\limits\_{n}a\_{n}$ ?


Thank you!



the limit of a scalar sequence in a sequence of scalars
Yes. Denote $B=\{b\_n\colon n=1,2,\ldots\}$. If $B$ is finite, then, since $\lim b\_n$ exists, the sequence $(b\_n)$ is eventually constant. If $B$ is infinite, there exists a subsequence $n\_1<n\_2<\ldots$ such that all $b\_{n\_j}$ are distinct. Denote $b\_{n\_j}=a\_{m\_j}$. All $m\_j$ are distinct, thus $\lim\_j a\_{m\_j}\leqslant \limsup a\_n$.


