# real analysis – Prove that \$limsup a_nleq sup a_n\$

So here’s one question which I came across some time ago.

Given any sequence $$a_n$$ in $$mathbb{R}$$, I need to show that $$limsup a_nleqsup a_n$$.

My approach: As the sequence $$a_ninmathbb{R}$$, thus to talk about the $$limsup a_n$$, I assumed that $$a_n$$ is bounded. I defined
$$b_n:=sup{a_k:kgeq n}$$ i.e.
begin{align}b_1&=sup{a_1,a_2,…}\b_2&=sup{a_2,a_3,…}\b_3&=sup{a_3,a_4,…}\.&\.end{align}
Now, I can say that $$$$ is a decreasing sequence. Thus, we have, $$b_1geq b_2geq b_3geq …$$ and so on.

Now, like we had defined $$b_n$$, we get that, $$sup a_n=b_1$$.

I then did this:
$$sup b_1=sup{b_1,b_2,…}geqinf{b_1,b_2,…}impliessup a_ngeqlimsup a_n$$
The reason why I did this is because $$$$ is a decreasing sequence. But I am a bit doubtful if I can write $$sup b_1=sup{b_1,b_2,…}$$. Can anyone please verify this?

Also, is there any alternate way of proving this inequality? Any help would be very much appreciable.