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?

Printable View

- September 25th 2006, 01:44 PMMKLyonAnalysis 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? - September 25th 2006, 01:51 PMThePerfectHacker
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. - September 25th 2006, 02:53 PMPlato
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.