# proof of a theorem

• Mar 4th 2008, 06:23 PM
Carol
proof of a theorem
Theorem states:
Note that a (sub n) is a constant sequence.
Suppose a (sub n) = a for all n greater than or equal to 1.
Then lim a (sub n) as n approaches infinity = a.

I need to prove this.
• Mar 4th 2008, 06:27 PM
ThePerfectHacker
Quote:

Originally Posted by Carol
Theorem states:
Note that a (sub n) is a constant sequence.
Suppose a (sub n) = a for all n greater than or equal to 1.
Then lim a (sub n) as n approaches infinity = a.

I need to prove this.

Is it true that $\displaystyle |a_n - a|<\epsilon$?
• Mar 4th 2008, 06:36 PM
Jhevon
Quote:

Originally Posted by Carol
Theorem states:
Note that a (sub n) is a constant sequence.
Suppose a (sub n) = a for all n greater than or equal to 1.
Then lim a (sub n) as n approaches infinity = a.

I need to prove this.

you need to use the definition of the limit for a sequence to show that $\displaystyle \lim_{n \to \infty}a_n = a$

that is, you must show that: for every $\displaystyle \epsilon > 0$, there exists an $\displaystyle N \in \mathbb{N}$ such that $\displaystyle n > N$ implies $\displaystyle |a_n - a|< \epsilon$