# Real analysis: Limit superior proof

1. Jan 16, 2010

### kingwinner

1. The problem statement, all variables and given/known data
Definition:
Let (an) be a sequence of real numbers. Then we define
lim [sup{an: n≥k}] = lim sup an
k->∞

(note: sup{an: n≥k} = sup{ak,ak+1,ak+2,...} = bk
(bk) is itself a sequence of real numbers, indexed by k)

Theorem:
Let a=lim sup an.
Then for all ε>0, there exists N such that if n≥N, then an<a+ε

2. Relevant equations
N/A

3. The attempt at a solution
I am trying to prove the theorem.
Proof:
Let bk=sup{an: n≥k}.
By the definition of "a" as a limit of supremums (bk->a), we have that
for all ε>0, there exists an integer N such that if k≥N, then
|sup{an: n≥k} - a| = |bk -a|< ε
=> sup{an: n≥k} < a + ε
By definition of an upper bound, sup{an: n≥k} ≥ an if n≥k.
So the above shows that if k≥N and n≥k, then an< a+ ε.

Now I am stuck...how should I continue? (I need to prove that: for all ε>0, there exists N such that if n≥N, then an<a+ε. But from my work so far, I'm really feeling hopeless)

There are so many different subscripts (e.g. n and k) to keep track of that I am really puzzled and frustrated now.

Can someone please help me out and show me the correct way to prove the theorem?
Any help is very much appreciated!

Last edited: Jan 16, 2010