Happy new year! I wish to prove the following result to practice with $\limsup: $
For a convergent sequence$ (b_n)$ we know $\limsup b_n = \lim b_n$
If $b_n$ is convergent, the set of accumulation points is definitely nonempty as it must contain the limit of $b_n$ and we know that $b_n$ is bounded. Therefore if $V$ is the set of accumulation points of $(b_n)$, we know that $\limsup(b_n)=\sup(V)$.
Intuitively and by previous exercises I know that since limits are unique, there must be precisely one accumulation point and therefore $V$ contains one element, namely $\lim(b_n)$. We now conclude that: $$\limsup b_n = \lim b_n $$
Was this argument convincing? Was I a bit sloppy, can I make this argument more precise, is it correct?
After this question I will attempt to prove a similar result for a divergent sequence $(b_n)\to \infty$ then $\limsup b_n = \infty$.
Edit/addition: $\limsup b_n = \infty$ follows immediately from unboundedness (from above) of a divergent sequence and the definition of $\limsup$, which is nice.