epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

164 The Nuts and Bolts of Proof, Third Edition
a. The smallest number we can use is fc = 2. We have to add fractions
whose denominators are integer numbers between 3 (which
corresponds to fc +1) and 4 (which corresponds to 2 k). Therefore,
the left-hand side of the equation becomes
3^4~l2*
Because 7/12 > 1/2, the statement is true in this case.
b. We assume that the inequality is true for an arbitrary number n.
Thus,
1 1 1 1
- + 7—TT—r + • • • + :^ > :^ •
n+1 (n+l) + l 2n 2'
c. We need to prove that the inequahty holds true for n+ 1. We will
add fractions with denominators between (?z + 1)+ 1 and 2(n+ 1).
So, we want to prove that
1 1 1 1 1
• + 7 7^—^+•••+:; 7 + w: 7T>:^.
(n+l) + l (n+l) + 2 2n-\-l 2(n-\-l) 2"
One thing to notice is that the largest denominator of the fractions in
the inductive hypothesis is n + 1, while the largest denominator in this
step is n + 2. Thus, to make the inequality in the inductive hypothesis
and the left-hand side of the inequality to be proved start with
fractions having the same denominator, we could rewrite the
inductive hypothesis as:
1 1 1 1 1
n + 2 (n + 2) + 1 2n 2 n + 1*
Using the associative property of addition and the rewritten inductive
hypothesis, we obtain
1 1 1 1
(n + 1) + 1 (n + 1) + 2 2n + 1 2(n + 1)
_ r 1 1 n 1 1
~ [('^ + 1) + 1 "^ (n + 1) + 2 "^ * * "^ 2nJ "^ 2n + 1 "^ 2(n + 1) -
/I 1 \ 1 1
^ V2 n + ly "^ 2n + 1 "^ 2(n + H 1)
_1 1 1
~2"^(2n+l)(2n + 2)^2'