Help with Sup and Inf of A = {(-1)^n + 1/n | n is a natural}

[daon]

I am given a set A = {(-1)^ⁿ + 1/n | n is a natural number}

I've done a few of these problems, but none so difficult as this. I need to find supA and infA and prove my answer.

Just plugging in numbers, I see the follwing pattern: 0,3/2,-2/3,5/4,-4/5,7/6,-6/7,9/8,-8/9

If I had to guess at this point, I'd say supA = 3/2 and infA = -1.

To show 3/2 is the greatest upper bound, I must show:
1) 3/2 is an u.b. and
2) if a is an u.b. of A then 3/2 < a.
OR
2') if a >= 3/2 there is an c in A s.t. c>a

Similarly for InfA, I need to show:
1) -1 is a l.b. for A
2) if a is a l.b. for A then -1 > a
OR
2') if -1 <= a then there is a c in A s.t. c<a

So I know what I have to do, I just don't know where to start. We are assuming to know all elementary properties of the real numbers.

Thanks,
-Daon

 

[stapel]

Random thoughts:

For the supremum, you'll be working with even values of n; that is, n = 2m for some natural number m. Then:

. . . . .(-1)ⁿ + 1/n = ((-1)²)^m + 1/(2m) = 1 + 1/(2m)

For all naturals j and k, j > k, we have 1/j < 1/k. Since 2m < 2(m + 1), then 1/(2m) > 1/(2(m + 1)). Then the maximal element in the set {1 + 1/(2m)} must occur at the least value of m, namely, m = 1.

For the infimum, you have n odd; that is, you have n = 2m - 1 for some natural m. Then:

. . . . .(-1)ⁿ + 1/n = ((-1)^2m)(-1)^-1 + 1/(2m - 1) = 1/(2m - 1) - 1

Following similar reasoning as above, 1/(2(m + 1) - 1) < 1/(2m - 1). Since you are adding the positive 1/n to the negative -1, the "lowest" value will occur at the smallest value of 1/n.

Hope that helps a bit.

Eliz.

 

[daon]

Thank you for the response. I'm not sure if I can distinguish between odd and even number elts in my proof. It may just be that I'm afraid to since nothing was discussed about it, but what you did would definetly simplify the problem a bit.

Anywho, I think I got the 1's on both parts just fiddeling with numbers, but I may have made some faulty assumptions. I also used some of what you said, stapel.

1) SupA: I showed this by induction:
3/2 >= (-1)^n + 1/n
For n=1 we get 3/2 >= 0.
Assuming P(n) is true: (-1)ⁿ⁺¹ + 1/(n+1) to show P(n+1) is true.

Since (-1)ⁿ⁺¹ <= 1, (-1)ⁿ⁺¹ + 1/(n+1) <= 1 + 1/(n+1) .

Also, since n>1, (here is where I utilize the inductive property, although I never use the hypothesis, is this OK?)
1/(n+1) <= 1/(1+1) = 1/2 so, 1/(n+1) <= 1/2, so:

1+1/(n+1) <= 1+1/2 = 3/2
THUS, (-1)ⁿ⁺¹ + 1/(n+1) < 3/2.

2) InfA
We know -1 <= (-1)ⁿ can I do this?
So, obviously -1 <= (-1)ⁿ + 1/n, since 1/n > 0 for all n.

Thanks

 

[pka]

You are making much too much out of 3/2 being LUB.
(3/2) is in the set and is an upper bound for set so it is the LUB.
That last property is easy to show.

You have shown that -1 is a lower bound.
Now showing -1 is the GLB is another matter altogether.
Suppose that \(\displaystyle \varepsilon > 0 ,\) then \(\displaystyle \left( {\exists n \in Z^ + } \right)\left[ {\frac{1}{n} < \varepsilon } \right].\)
The odd number \(\displaystyle 2n + 1\) is such that \(\displaystyle 2n + 1 > n\, \Rightarrow \;\frac{1}{{2n + 1}} < \frac{1}{n} < \varepsilon .\)
Therefore for each \(\displaystyle \varepsilon > 0\) there is an element in the set such that \(\displaystyle - 1 < - 1 + \frac{1}{{2n + 1}} < - 1 + \varepsilon .\) Now that is sufficient to prove that -1 is GLB.

 

[daon]

pka, I'm not too sure why you used the fact that 2n+1>n, but I made a guess. I am still having issues with "reading into" mathematics, so forgive me if it is obvious.

This is what I think you meant, let me know if it requires modification:

Suppose that there is a lower bound a s.t. a>-1. Then a = -1+e for some e in R+.
But by 1/n principle, there exists a natural number n s.t. 1/n < e. So let k be the first odd natural number >= n s.t. 1/n < e. So, 1/k < e.

Notice that (-1)^k + 1/k is the kth element in A, and that (-1)^k = -1 since k=2l+1 for some l and (-1)^k=(-1)^(2l)(-1) = -1.
Thus (-1)^k + 1/k = -1 + 1/k <= -1 + e = a. But -1+1/k is an element in A, so a cannot be a lower bound (contradiction).

edit: Pka, your signature was just on Jeopardy under the "math joke" category :lol:

 

[pka]

pka, I'm not too sure why you used the fact that 2n+1>n.

That is easy to answer. For any integer n, 2n+1 is an odd integer greater than n.
The sequence term s<SUB>2n+1</SUB>=(-1)<SUP>2n+1</SUP>+1/(2n+1)= -1+1/(2n+1).
Thus for any number greater than –1 there is a term of the sequence between that number and (-1). That mean that –1 is the GLB.