Finding the Supremum of S: A Proof

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SUMMARY

The supremum of the set S defined as S:={1-(-1)^n /n: n in N} is conclusively determined to be 2. The proof establishes that 2 is an upper bound for all elements in S, as 1-(-1)^n /n is always less than or equal to 2 for any natural number n. Furthermore, it is demonstrated that any upper bound v less than 2 cannot be valid, as it fails to satisfy the condition for all n in N. Therefore, supS=2 is confirmed as the least upper bound.

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Homework Statement



Let S:={1-(-1)^n /n: n in N}
Find supS.

The Attempt at a Solution



Is this the right way to write the solution?
Thanks!

First, I want to show that 2 is an upper bound.
For any n in N, 1-(-1)^n /n is less or equal to 2. Thus, n=2 is an upper bound.

Second, I want to show that 2 is the least upper bound.
Assume there is an arbitrary upper bound v, such that v<2. Then, however, v< 1-(-1)^n /n for any n in N. Thus, v is not an upper bound. Hence, supS=2.
 
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Kinetica said:

Homework Statement



Let S:={1-(-1)^n /n: n in N}
Find supS.


The Attempt at a Solution



Is this the right way to write the solution?
Thanks!

First, I want to show that 2 is an upper bound.
For any n in N, 1-(-1)^n /n is less or equal to 2. Thus, n=2 is an upper bound.

Why is 1-(-1)^n/n always less or equal than 2?

Second, I want to show that 2 is the least upper bound.
Assume there is an arbitrary upper bound v, such that v<2. Then, however, v< 1-(-1)^n /n for any n in N. Thus, v is not an upper bound. Hence, supS=2.

Why is v < 1-(-1)^n/n for any n in N?? (this is not true) You only need v<1-(-1)^n for one n in N. Which one?
 
Pretty close, hopefully the mistake was just a typo:

"Then, however, v< 1-(-1)^n /n for any n in N."

Replace "any" with "some" and it's fine. Can you see why?

EDIT: Too slow lol.
 

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