Every number in that decreasing sequence is less than or equal than 1 so it is an upper bound. Since it is in the set itself, no upper bound can be less than 1 so 1 is the least upper bound- the supremum. (If an upper bound for a set is in the set, if it is the largest number in the set, or maximum, then it is the least upper bound= supremum.
The "infimum" of a set of real numbers is their 'greatest lower bound'.
There is no smallest number in the set- they just keep getting smaller as you go out the sequenc. Every number in this set is positive so 0 is a lower bound. But it should be clear that the limit of the sequence is 0. If the limit of a sequence is also a lower bound, then it is the greatest lower bound= infimum. That's true because if we claimed that some number, , greater than the limit, were a lower bound, we could take the " " in the definition of limit to be where L is the limit of the sequence. Since there must be numbers in the sequence closer to L than , there must be numbers in the sequence less than - it is NOT a lower bound.
Although it is not necessary here, if the limit of a sequence is also an upper bound, then it is the least upper bound= supermum.
For every decreasing sequence, then, the first number in the sequence is the supremum and the limit of the sequence is the infimum.
For every increasing sequence, the first number in the sequence is the infimum and the limit of the sequence is the supremum.