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?
We need to show,
lim n--> oo a_n=0
For all e>0 there is an N such that,
if n>N then, |a_n|=a_n<e
We note that,
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
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.