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

170 The Nuts and Bolts of Proof, Third Edition
n^ -\-n is an even number. If n is an odd number, then n = 2/c +1
for some integer number k. Thus,
n^ + n = (2/c + l)[(2/c + 1) + 1]
= (2/c + l)(2/c + 2) = 2[(2/c + l)(/c + 1)].
Because the number (2/c+l)(/c+1) is an integer, this proves that
n^ + n is an even number. Therefore, the statement is true.
39. We will prove this statement by induction.
a. Let us check if the inequality holds for /c = 6. In this case, we obtain
6! = 720>216=-6l
b. Assume that n\>n^.
c. We have to prove that (n+l)!>(n+1)^. By the properties of
factorials, {n + 1)! = (n + l)n!. If we use this fact and the inductive
hypothesis, we obtain {n -f 1)! — {n-\- \)n\ > (n + l)n^. We will now
use this inequality and other algebraic properties of inequalities
to obtain the expression n^ + 3n^ + 3n + 1 = (n + 1)^:
(n + 1)! = (n + \)n\ >{n+ \)n^
> (6 + l)n^ = n^ + 6n^
> n^ + 6n^ = n^ -\- 3n^ + 3n^
= n^ + 3n^ + 3n X n > n^ + 3n^ + 3 x 6 x n
> n^ + 3n^ + 3n + n > n^ + 3n^ + 3n + 1
= (n^l)\
Therefore, by the principle of mathematical induction, the original
statement is true for all integers k> 6.
40. Assume that lim„^oo{(-l)"3} = L, where Lis a real number. Then,
for every £>0, there exists an iV > 0 such that |(-1)"^ - L| <e for
all n>N. This implies that -e<L-(-l)"(l/5)<6:; that is,
-e + (-l)"(l/5)<L<f+ (-!)"3. Because these inequalities will
hold true for all n> N, they will hold true for odd and even
values of n. Thus, one has to consider the following two sets of
inequahties:
1 , 1
-s + -<L<£-\--