Page 373 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 359
Then
n+1 n
1 1 1 1 1
= + ≤ 2 − +
i 2 i 2 (n + 1) 2 n (n + 1) 2
i=1 i=1
2
2
n + n + 1 n + n 1
= 2 − 2 < 2 − 2 = 2 − .
n(n + 1) n(n + 1) n + 1
8. (a) We let m be arbitrary and then prove by induction that for all n ≥ m,
H n − H m ≥ (n − m)/n.
Base case: n = m. Then H n − H m = 0 ≥ 0 = (n − m)/n.
Induction step: Suppose that n ≥ m and H n − H m ≥ (n − m)/n.
Then
1 n − m 1
H n+1 − H m = H n + − H m ≥ +
n + 1 n n + 1
2
2
n + n − mn − m + n n + n − mn
= ≥
n(n + 1) n(n + 1)
n + 1 − m
= .
n + 1
(b) Base case: If n = 0 then H 2 = H 1 = 1 ≥ 1 = 1 + n/2.
n
Induction step: Suppose n ≥ 0 and H 2 ≥ 1 + n/2. By part (a),
n
2 n+1 − 2 n 1
H 2 n+1 − H 2 ≥ n+1 = .
n
2 2
Therefore
1 n 1 n + 1
n
H 2 n+1 ≥ H 2 + ≥ 1 + + = 1 + .
2 2 2 2
(c) Since lim n→∞ 1 + n/2 =∞, by part (b) lim n→∞ H 2 =∞. Clearly
n
the H n ’s form an increasing sequence, so lim n→∞ H n =∞.
n
12. (a) Hint: Try proving that 2 ≥ n + 1, from which the desired conclusion
follows.
n 2
(b) Base case: n = 9. Then n! = 362880 ≥ 262144 = (2 ) .
n 2
Induction step: Suppose that n ≥ 9 and n! ≥ (2 ) . Then
2
n 2
(n + 1)! = (n + 1) · n! ≥ (n + 1) · (2 ) ≥ 10 · 2 2n ≥ 2 · 2 2n
= 2 2n+2 = (2 n+1 2
) .
2
(c) Base case: n = 0. Then n! = 1 ≤ 1 = 2 (n ) .
2
Induction step: Suppose that n! ≤ 2 (n ) . Then
2
2
2
2
2 ((n+1) ) = 2 n +2n+1 = 2 (n ) · 2 2n+1 ≥ 2 (n ) · 2 n+1 > n! · (n + 1)
= (n + 1)!