Test Review 4 - is it this easy?

[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

 

[HallsofIvy]

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

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

 

[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.

 

[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}