Page 119 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

80
1
3.87 For k ∈ N , show 16 < √ < 17.
k
k=1
√ √ √ √ √ √ √ √
Proof: From k − 1 < k< k +1, we have k + k − 1 < 2 k< k +1+ k .
k ∈ N , then
√ √ √ √ √ √
1 √ √ 1 √ √ 1√ √
√ √ < 12 k< √ 1 √ ⇒ 2( k +1 − k) < √ < 2( k − k − 1) ⇒
1
1
k+1+ k √ < 12 k< √ k+ k−1 ⇒ 2( k +1 − k) < √ < 2( k k − k − 1) ⇒
√
√
k+1+ k n k+ k−1 k
n
√ √ √ √ 1 1 √ √ √ √ , where 1 ≤ m ≤ n , and m, n ∈ N . Choose

2( n +1
m)
√ < 2( n −m − 1)
2( n +1 − − m) < < √ < 2( n − m − 1)
k k
k=m 80
k=m
1
n = 80,m =1, then 16 < √ . Choose n = 80,m =2, then
k
80 80 k=1 80
1 1 √ √ √ √ 1

1+
1+ √ < 2( 80 − 1) + 1 < 2 81 − 1 = 17 17. Hence, 16 < √ < 17.
√ < 2( 80 − 1) + 1 < 2 81 − 1 =

k k k
k=2
k=2 k=1
) .
3.88 For n ∈ N ,n > 1, show n! < ( n+1 n
2
Proof 1: (applying mean inequality)
n!=1 · 2 · ··· · k · ··· · (n − 1) · n (i),
n!= n · (n − 1) · ··· · (n − k + 1) · ··· · 2 · 1 (ii).
2
) =(
) =
(i)× (ii): (n!) = (1 · n)[2(n − 1)] · ··· · [k(n − k + 1)] · ··· · [(n − 1)2](n · 1). Since 1 · n ≤ ( 1+n 2 2+n−1 2 1+n 2 k+n−k+1 2
) , 2(n − 1) ≤ (
) , ··· ,k(n − k + 1) ≤ (
2 2
2
2
2 2
1+n 2 ,k(n − k + 1) ≤ (
) =
)
1
n−1+2 2 (
1+n 2 · n ≤ ( 1+n ) , 2(n − 1) ≤ 2+n−1 2 1+n ) , ··· k+n−k+1 2
1+n 2 =(
( ) , ··· , (n − 1)2 ≤ ( ) =( ) ,n · 1 ≤ ( ) 2
2
2
2
2
2 2
2 2
2 2
) =(
) , ··· ,k(n − k + 1) ≤ (
1 · n ≤ ( 1+n 2 2+n−1 2 1+n 2 k+n−k+1 2 ( 1+n ) , ··· , (n − 1)2 ≤ ( n−1+2 2 1+n ) ,n · 1 ≤ ( 1+n )
) , 2(n − 1) ≤ (
) =(
) =

2−1 2
2n 2
2
2n 2
)
n−1+2 2(
1+n 2 ,k(n − k + 1) ≤ (
) =
1+n 2 · n ≤ ( 1+ ) , 2(n − 1) ≤ 2+n 1+n 2=( 1+ ) , ··· k+n−k+1 2 2 2 2 2
1
( ) , ··· , (n − 1)2 ≤ ( ) =( 2 ) ,n · 1 ≤ ( ) 2
2
2
2
2
2n 2
2n 2
) ] ⇔ (n!) ≤ (
)
) . There are n terms. Thus (n!) ≤ [(
)
) =(
( 1+ ) , ··· , (n − 1)2 ≤ ( n−1+2 2 1+ ) ,n · 1 ≤ ( 1+n 2 2 1+n 2 n 2 n+1 2n ⇔ n! ≤ ( n+1 n
2 2 2 2 2 2 2
2
) .
) . Since n> 1, then n! < (
) ] ⇔ (n!) ≤ (
(n!) ≤ [( 1+n 2 n 2 n+1 2n ⇔ n! ≤ ( n+1 n n+1 n
)
2 2 2 2
Proof 2: (applying mathematical induction)
) = . The inequality holds.
When n =2, LHS=2! =2, RHS=( 2+1 2 9
2 4
) . Then
Assume the inequality holds for n = k , that is, k! < ( k+1 k
2
k+1
k+1
) (k + 1)) (k + 1).
k!(k + 1) < (k!(k + 1) < ( k+1 kk k+1 kk
) (k + 1) ⇔ (k + 1)! < () (k + 1) ⇔ (k + 1)! < (
2 2 2 2
k+1 k k+1
k+1 k+1
(k+1)+1 k+1
( k+1 k (k+1)+1 k+1 ⇒ 2( k+1 k ) < [ (k+1)+1 k+1 ⇒ 2( k+1 k k+1 (k+1)+1 k+1 ⇒ k+1 k+1 (k+1)+1 k+1
< [
]
) (k+1) < [
) (
]
]
)
(k+1)+1 k+1
2 2 ( 2 ) (k+1) < [ 2 ] ⇒ 2( 2 ) ( ) < 2[ ] ⇒ 2( ) < [ ] ⇒
2
(k+1)+1 k+1 2 ) k+1 2 2 2 2 2 2 2
]
( (k+1)+1 k+1
]
]
⇒ 2(
2 )
) (
]
) (k+1) < [
( k+1 k (k+1)+1 k+1 ⇒ 2( k+1 k k+1 ) < [ (k+1)+1 k+1 2 < [ k+1 k+1 < [ k+1 ] ⇒ 2 < [ (k+1)+1 k+1 ( 2 ) k+1 . Binomial theorem implies that
2 2 2 2 2 2 2 2 k+1
]
2 < [ (k+1)+1 k+1 ( 2 ) k+1 1 k+1 1 k+1 kk (k+1)+1 k+1k+1
(k+1)+1
k+1
) (k + 1)
2 k+1 (1 + ) = 1 +(k + 1) + ··· > 2. Thus (( ) (k + 1) < [ < [ ]] holds.
k+1 k+1 2 2 2 2
) holds according to mathematical induction
Hence, the original inequality n! < ( n+1 n
2
above.
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