Question regarding infima and suprema proof

I'm trying to answer questions regarding infima and suprema but I am not sure on how to answer this question. It goes like this:

Let $\displaystyle A=${$\displaystyle \frac 12, \frac34, \frac 78, \cdots $}. Prove that A is unbounded and that sup(A)=1.

I have rewritten this as:

$\displaystyle A=${$\displaystyle 1-\frac 12, 1-\frac14, 1-\frac 18, \cdots $}

So that the n-th term is $\displaystyle 1-\frac{1}{2^n}$

I am not sure if I needed to find the n-th term but I am not sure what I can do with this question?

EDIT I have solved this question now so no need for an answer. Thanks.

Re: Question regarding infima and suprema proof

Quote:

Originally Posted by

**MichaelH** I'm trying to answer questions regarding infima and suprema but I am not sure on how to answer this question. It goes like this:

Let $\displaystyle A=${$\displaystyle \frac 12, \frac34, \frac 78, \cdots $}. Prove that A is unbounded and that sup(A)=1.

I have rewritten this as:

$\displaystyle A=${$\displaystyle 1-\frac 12, 1-\frac14, 1-\frac 18, \cdots $}

So that the n-th term is $\displaystyle 1-\frac{1}{2^n}$

I am not sure if I needed to find the n-th term but I am not sure what I can do with this question?

EDIT I have solved this question now so no need for an answer. Thanks.

do you mean prove it's bounded? an unbounded set or sequence has no supremum.

From your form of A it's pretty clear A is bounded by 1. Just prove that B=1 is the smallest number that satisfies B > A_{n} for all n

Re: Question regarding infima and suprema proof

To prove that "A is bounded" (as opposed to "bounded above") you need to show also that there is a lower bound. Of course, here, every number is positive so 0 is an obvious lower bound.