# Supremums and infimums

1. Dec 27, 2008

### Fairy111

1. The problem statement, all variables and given/known data

In the following case decide, whether the set S has a supremum or an infimum.

2. Relevant equations

S= {1+(-1)^n times (1/n), n is part of the natural numbers \ {0}}

3. The attempt at a solution

I first started to form the set S but substituting in numbers for n,

so i came up with the set, { 1, 0, 3/2, 2/3 ...}

So i thought that the infimum would be 1, as that is the largest lower bound. Im not sure what the supremum is though.

2. Dec 27, 2008

### mutton

But 2/3 < 1.

3. Dec 27, 2008

### Fairy111

sorry, yes 2/3 is the infimum. How do i find the supremum though?

4. Dec 27, 2008

### snipez90

The definition of your set seems a bit weird. Are you saying that n is a natural number (typically denoting a positive integer) not including 0? The formula wouldn't be defined for n = 0, so I will assume that we start at n = 1. For n = 1, you get 1 + (-1) = 0; for n = 3, you get 2/3; for n = 5, you get 4/5; for n = 7, you get 6/7. In general, for n = 2k + 1, where k is a nonnegative integer, you will get 1 + (-1)^(2k+1) * [1/(2k+1)] = 1 - [1/(2k+1)]. It should be clear what happens as k gets larger. This handles odd n (remember that 0 is in the set).

After analyzing the even terms on your own, it should be clear what the sup and inf are.

5. Dec 27, 2008

### HallsofIvy

Staff Emeritus
No, that's incorrect. You may be trying to rush this because you are not even using the formula correctly. The first number, n= 1, is (1- (-1)1)/1= 0. Then n= 2 gives (1- (-1))2)/2= 2/2= 1, not 3/2. n= 3 gives (1-(1)3)/3= 0/3= 0. n= 4 gives (1- (-1)4)/4= 2/4= 1/2. It should be obvious that for n any odd number, (-1)n= -1 so (1+ (-1)n)/n= 0/n= 0. For any even n, (-1)n= 1 so (1+ (-1)n)/n= 2/n.

You need to review the definitions! The infimum of a set is the "largest of all lower bounds", that is correct. But a "lower bound" is a number less than or equal to any number in the set. "1" is not a lower bound at all because 0< 1 so 1 is not "less than or equal to any number in the set".
Similarly, 2/3 is NOT the infimum because, again, it is not a lower bound at all: 0< 2/3.

Every number in this set is either 0 or 2/n. It should be easy to see what the infimum is from that! and, since 2/n< 1 for all n> 2 it should also be easy to see what the supremum is. Notice that, for this set, both infimum and supremum are in the set. That does not always happen.

I notice that you are not actually required to find the supremum and infimum, only decide whether or not they exist. But finding something is the best way to show it exists!

Last edited: Dec 27, 2008
6. Dec 27, 2008

### snipez90

eh, I think problem is slightly more interesting if the formula was 1 + [(-1)^n / n] but hallsofivy might have the more plausible formula.