How to prove an increasing sequence that converges is bounded above by its limit I am trying to prove that an increasing sequence that converges to $ L$ is bounded above by its limit.
By using  $a_n \le a_{n+1}$ and the definition of limit of a sequence, I can prove that for $\epsilon > 0$ , $ a_n \lt {L + \epsilon} $ for all $a_n$.
But is there a way to proceed to $ a_n \le L $ ? because I can't think of a case in which the former is true but the latter isn't.
 A: HINT
You can easily show that if for some n $a_n>L$ then by definition of limit $a_n$ must decrease which is impossible.
You only need to formalize this idea by setting “assume exists n such that ...then by definition of limit...contradiction”.
Notably 


*

*suppose $\exists n_1$ such that $a_{n_1}>L$ with $d=a_{n_1}-L>0$ 

*set $\epsilon=d$ by definition of limit must exists $n_2>n_1$ such that $|a_{n_2} -L|<\epsilon \implies a_{n_2}<a_{n_1}$

A: If $a_n$ converges to $L$ and is increasing then let $\epsilon>0 \ ,\exists N \in \Bbb N \ s.t. \ \forall n>N \ \  \vert a_n-L\vert < \epsilon$
We can open up this up as you did to $a_n<L+\epsilon$
As indicated in the hint above, if $\exists n $ such that $L<a_n<L+\epsilon$ we get a contradiction because the limit is $L$ and so the sequence will have to "decrease" from $L<a_n<L+\epsilon$ down to $L$ 
A: If a term say,  $a_n$ gets larger than the limit $L$, then all the other terms, $a_{n+1}, a_{n+2},...$  must stay above $a_n,$ since your sequence is increasing.  
Thus for $$\epsilon = \frac {a_n-L}{2}$$ all the terms $a_k$ for $ k\ge n$  stay out of the neighbourhood, $$(L-\epsilon, L+\epsilon)$$ which contradicts the assumption of  $$lim _{n\to \infty} a_n = L.$$  
