The Real And Complex Number Systems

(l) ∑ n2
n p 1
n−1 − 1 n
Proof: Note that
n p 1
n − 1 −
1 n
1
n 3 2 −p
n
n − 1
1
1
n−1
n
So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit
Comparison Test.
(m) ∑ n1
n 1/n − 1 n
Proof: With help of Root Test,
lim n→
sup n 1/n − 1 n 1/n 0 1,
the series converges.
(n) ∑ n1
n p n 1 − 2 n n − 1
Proof: Note that
n p n 1 − 2 n n − 1
1
n 3 2
n 3 2 −p n n 1 n n − 1 n − 1 n 1
.
So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit
Comparison Test.
8.16 Let S n 1 , n 2 ,... denote the collection of those positive integers that do not
involve the digit 0 is their decimal representation. (For example, 7 ∈ S but 101 ∉ S.)
Show that ∑ k1
1/n k converges and has a sum less than 90.
Proof: DefineS j the j − digit number ⊆ S. Then#S j 9 j and S
j1 S j . Note
that
∑ 1/n k 9 j
10 . j−1 k∈S j
So,
∑
k1
1/n k ≤ ∑
j1
9 j
10 j−1 90.
In addition, it is easy to know that ∑ k1
1/n k ≠ 90. Hence, we have proved that ∑ k1
1/n k
converges and has a sum less than 90.
8.17 Given integers a 1 , a 2 ,...such that 1 ≤ a n ≤ n − 1, n 2, 3, . . . Show that the
sum of the series ∑ n1
a n /n! is rational if and only if there exists an integer N such that
a n n − 1 for all n ≥ N. Hint: For sufficency, show that ∑
n2
n − 1/n! is a telescoping
series with sum 1.
Proof: ()Assume that there exists an integer N such that a n n − 1 for all n ≥ N.
Then
.