Need help proving sequences involving inferiors $$L = \inf\limits_{n\ge1}   b_n  $$ where $$b_n = \sup\limits _{k\ge n} a_k $$
and $$ \{a_k\}_ {k =1}^{\infty}$$
I need to show that if $\epsilon$ is any positive number, then there is an integer N such that $\lvert b_n - L \rvert < \epsilon  $ for all $n \ge N$ 
Please help!
 A: Let $\varepsilon>0$. By definition of $L$, there is an $N\in\mathbb N$ s.t. $$L\leq b_N<L+\varepsilon.$$
Therefore, since $(b_n)$ is decreasing, for all $n>N$,
$$L\leq b_n\leq b_N<L+\varepsilon\implies |b_n-L|<\varepsilon.$$
A: First you should suppose that $b_1=\sup\limits_{k\ge 1}{a_k}<\infty$, i.e $b_1$ is finite. Then because for each next $b_n$ you take supremum over a smaller set of $a_k$ means that $b_n$ is a monotonically decreasing sequence of numbers. By assumption you know that it is bounded from below by the finite number $L$ which is also its infimum, i.e $L=\inf\limits_{n\ge 1}{b_n}$. In this way we conclude that the sequence $\{b_n\}_{n=1}^\infty$ has a limit point. Now we prove that this limit is $L$. 
By assumption we have that $L=\inf\limits_{n\ge 1}{b_n}$. This means that for each $\epsilon>0$ there exists some element $b_{n_0}$ such that $L\leq b_{n_0}<L+\epsilon$. But also we showed that $b_n$ is a monotonically decreasing sequence, so $b_{n_0} \ge b_n,\,\,\forall n>n_0:$ which means $L\leq b_n\leq b_{n_0}<L+\epsilon,\,\,\forall n>n_0$. The last is equivalent to $|L-b_n|<\epsilon,\,\,\forall n>n_0=:N$
