Let

. You should note first that the sequence

is decreasing hence

.

Since

, you must prove:

a) that

for any convergent subsequence

b) that, for any

, there is a convergent subsequence

such that

.

I let you try a) (compare

with terms from the sequence

)

As for b), you have to "construct" a convergent subsequence. In fact, instead of using

like above, you can directly construct a subsequence that converges toward

(which means that in fact

). To do so, intuitively, choose

close to

(it is a sup, hence it is a limit of values

in

), then

close(r) to

, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)

NB: you'll probably have to distinguish cases depending whether

is finite or not.

--

About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.