# Test Review 4 - is it this easy?

1. Oct 15, 2005

### cmurphy

Let (sn) be a real sequence and s in R. Prove that lim sup (sn) < s implies
sn < s for n large.

My answer seems too easy. Is there anything missing?

Given lim sup (sn) < s.
By definition of lim sup, we know lim N->infinity sup {sn: n > N} < s
Then for n > N, we must have sn < s.

Colleen

2. Oct 15, 2005

### HallsofIvy

Staff Emeritus
What exactly is the "definition of lim sup"? How does
lim N->infinity sup {sn: n > N} < s follow from it?

3. Oct 15, 2005

### cmurphy

The definition in our books is that:

lim sup {sn: n>N}
n->inf

i.e. For large n, the lim sup sn is the limit of all of the suprema. They also defined lim sup sn to be exactly the supremum of the set of subsequential limits.

4. Oct 15, 2005

### cmurphy

Also, the fact that lim sup sn < s for some real number s is given.

Then I just made the substitution lim sup sn = lim n->infinity sup{sn: n>N}