Prove: $\limsup_{n\to\infty} (a_n)\leq \sup(a_n)_{n=1}^\infty$

Question

Proposition: Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers. Show that $\limsup_{n\to\infty} (a_n)\leq \sup(a_n)_{n=1}^\infty$.

Any hints on how to prove it will be appreciated.

Answer 1

It's pretty clear: $\,\,\limsup_{n\to\infty}a_n$ is the limit of the nonincreasing sequence in $\overline{\mathbf R} $: $\sup_{k\geq1}a_k,\,\,\sup_{k\geq 2}a_k, \dots,\,\,\sup_{k\geq n} a_k, \dots $

So, fot all $n$, we have $\sup_{k\geq n} a_k\le \sup_{k\geq1}a_k$. There remains to pass to the limit.

Answer 2

Let me give you a hint to start:

Case $1$: If $\sup (a_n)=\infty$, we have nothing to do.

Case $2$: now suppose $\sup_n (a_n)<\infty$, think about the definition of $\limsup (a_n)$. We have $$\limsup_{n\to\infty} a_n:=\lim_{k\to\infty} (\sup_{m\geq k}a_m) $$ Clearly, $$ \sup_{m\geq k}a_m\leq \sup_n (a_n) $$ for arbitrary integer $k>0$.

Now, could you take it from here?