# Analysis Proof

• Sep 25th 2006, 12:44 PM
MKLyon
Analysis Proof
Let E be greater than zero. Show that there is a n in N (natural numbers) such that 1/(2^n) is less than E.

How would you go about proving this?
• Sep 25th 2006, 12:51 PM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
Let E be greater than zero. Show that there is a n in N (natural numbers) such that 1/(2^n) is less than E.

How would you go about proving this?

Consider the sequence,
a_n={1/2^n}
We need to show,
lim n--> oo a_n=0
By definition,
For all e>0 there is an N such that,
if n>N then, |a_n|=a_n<e

We note that,
0<=a_n<=1/n
And the limits of the sequenes,
lim n---> oo 0=0
lim n---> oo 1/n=0
Thus, by the Squeeze theorem for sequnces we have,
lim n---> oo a_n=0
Q.E.D.
• Sep 25th 2006, 01:53 PM
Plato
Here is a second way.

Lemma: If E>0 there is a positive integer n such that (1/n)<E.
Proof: (1/E)>0 and the positive integers are not bounded above.
Therefore, the is an integer n such that n>(1/E) or (1/n)<E.

Now note that for each n, n<2^n. So (1/2^n)<(1/n)<E.