Solved Problems on Supremum and Infimum

1
Solved Problems on Supremum and Infimum
1
2
Sup and Inf
Problem 1
Solution
Use the Maple commands gt A unapply((n2(-1)n
n1)/(n2(-1)(n1)n2),n)gt for k to 10 do
evalf(A(k))od to list the first ten elements of
the sequence A.
Floating point approximations of the elements
are 0.250, 1.750, 0.500, 1.500, 0.656, 1.343,
0.741, 1.258, 0.793, 1.206.
It appears that the odd elements form an
increasing sequence and the even elements a
decreasing sequence. Furthermore, the odd
elements appear to be smaller than the even
elements.
3
Sup and Inf
Problem 1
Solution (contd)
Assuming that the previous observation about the
odd and even elements is true, sup(A) 1¾ and
inf(A) ¼.
It remains to show that the odd elements of A
form an increasing sequence and the even elements
a decreasing sequence.

• Assuming that the function A is already defined
  for Maple, use the Maple commands
• gt assume(n,posint)
• gt simplify(A(2n2)-A(2n))
• to compute the difference of two subsequent even
  elements of the set A.

4
Sup and Inf
Problem 1
Solution (contd)
One gets
One concludes that A(2n2) A(2n) is always
negative.
Hence the even elements of the sequence A form
a decreasing sequence.
5
Sup and Inf
Problem 1
Solution (contd)
In the same way one gets
One concludes that A(2n 3) A(2n 1) is
always positive.
Hence the odd elements of the sequence A form a
increasing sequence.
6
Sup and Inf
Problem 1
Solution (contd)
Conclude
7
Sup and Inf
Problem 1
Solution (contd)
We have shown that the odd elements of the
sequence A form an increasing sequence with the
limit 1.
We have also shown that the even elements of the
sequence A form an decreasing sequence with the
limit 1.
Here odd elements are denoted by red diamonds and
even elements by blue.
Conclude sup(A) A(2) 7/4 and inf(A)
A(1)1/4.