#NinZZ# #(1/2)^(2n-1) < (1/2)^(n²-1)# Max(n) = ? a)1 b)2 c)3 d)4 e)5

2 Answers
Douglas K.
May 9, 2017

Selection a)1

Explanation:

The equality point is:

#2n -1= n^2-1#

#2n = n^2#

#2 = n#

Therefore, the inequality is true for #n < 2#.

That leaves only selection a)1

Cesareo R.
May 9, 2017

#0 < n < 2 rArr max(n) < 2#

Explanation:

#log# is a strictly increasing transformation so if #a < b# then

#log a < log b#

Making #a = (1/2)^(2n-1)# and #b = (1/2)^(n^2-1)# then

#(2n-1)log (1/2) < (n^2-1) log (1/2)# but #log(1/2) < 0# so the condition

#(1/2)^(2n-1) < (1/2)^(n^2-1)# reduces to

#2n-1> n^2-1#

or

#n^2 < 2n#

or

#n(n-2) < 0# or

#0 < n < 2#

NOTE:

Strictly speaking if #n_0# is the maximum then

1) #n_0 in (0,2)#
2) #forall n in (0,2) rArr n le n_0#

but this fails because for any #n in (0,2) rArr n < (n+2)/2 < 2# and then #n_0 < (n_0+2)/2 < 2# also! So #(0,2)# does not have maximum. But this open set has a supremum which is #2#.

In terms of sets, the maximum is the largest member of the set, while the supremum is the smallest upper bound of the set.

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