Solved Problems on
Supremum and Infimum
n 2 + ( −1)n n + 1
1 Let A = 2 n = 1,2,3,… .
( )
n +1
n + − 1 n + 2
Determine sup ( A ) and inf ( A ) .
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3,… .
n + ( −1) n + 2
n +1
Solution
Determine sup ( A ) and inf ( A ) .
Use the Maple commands
> A := unapply((n^2+(-1)^n*n+1)/(n^2+(-1)^(n+1)*n+2),n);
> 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.
Mika Seppälä: Solved Sup and Inf Problems
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3… .
n + ( −1) n + 2
n +1
Determine sup ( A ) and inf ( A ) .
Solution (cont’d)
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
> assume(n,posint); Maple needs to know this in order
> simplify(A(2*n+2)-A(2*n)); to be able to simplify correctly.
to compute the difference of two subsequent even elements of the set
A.
Mika Seppälä: Solved Sup and Inf Problems
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3,… .
n + ( −1) n + 2
n +1
Determine sup ( A ) and inf ( A ) .
Solution (cont’d)
One gets
2
8 n~ + 4 n~ − 5
A( 2 n + 2 ) − A( 2 n ) = −
2 2
2 ( 2 n~ + 3 n~ + 2 ) ( 2 n~ − n~ + 1 )
One concludes that A(2n+2) – A(2n) is always negative.
Hence the even elements of the sequence A form a decreasing
sequence.
Mika Seppälä: Solved Sup and Inf Problems
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3,… .
n + ( −1) n + 2
n +1
Determine sup ( A ) and inf ( A ) .
Solution (cont’d)
In the same way one gets
2
8 n~ + 20 n~ + 7
A( 2 n + 3 ) − A( 2 n + 1 ) =
2 2
2 ( 2 n~ + 7 n~ + 7 ) ( 2 n~ + 3 n~ + 2 )
One concludes that A(2n + 3) – A(2n + 1) is always positive.
Hence the odd elements of the sequence A form a increasing
sequence.
Mika Seppälä: Solved Sup and Inf Problems
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3,… .
n + ( −1) n + 2
n +1
Determine sup ( A ) and inf ( A ) .
n 2 + ( −1) n + 1
n
Solution (cont’d) Let A ( n ) = .
n + ( −1)
n +1
2
n+2
( −1)
n
1
n + ( −1) n + 1
2 n 1− 2
+
lim A ( n ) = lim 2 = lim n n = 1.
n + ( −1) n + 2 ( −1)
n →∞ n →∞ n +1 n →∞ n +1
2
1+ + 2
n n
Conclude
lim A ( 2n ) = lim A ( 2n + 1) = lim A ( n ) = 1.
n →∞ n →∞ n →∞
Mika Seppälä: Solved Sup and Inf Problems
Sup and Inf
n 2 + ( −1)n n + 1
Problem 1 Let A = 2 n = 1,2,3,… .
n + ( −1) n + 2
n +1
Determine sup ( A ) and inf ( A ) .
Solution (cont’d)
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.
Mika Seppälä: Solved Sup and Inf Problems
