(n+1)( 32 )n+1 2
7. Ratio Test: lim n 2 n = lim n+1
= 23 < 1. Hence absolutely convergent.
(3)
n→∞ n 3
n→∞
10n+1 2n+3 10
= 58 < 1. Hence absolutely convergent.
13. Ratio Test: lim (n+2)4
= lim n+1
10n
n→∞ (n+1)42n+1 n→∞ n+2 16
q
n2 +1 n
21. Root Test: lim n 2n
= lim
2 +1
n2 +1
2 +1
2n
n→∞
n→∞
=
1
2
< 1. Hence absolutely convergent.
q
2
n
1 + 1 n = lim (1 + 1 )n = e > 1. Hence divergent.
23. Root Test: lim
n
n
n→∞
n→∞
27. Define (2n − 1)!! = 1 · 3 · · · (2n − 1). Ratio Test:
(−1)n (2n+1)!! 2n + 1
(2n + 1)!! (2n − 1)!
(2n+1)! lim =
lim
=0
=
lim
n→∞ (−1)n−1 (2n−1)!! n→∞ (2n + 1)(2n)
n→∞ (2n − 1)!! (2n + 1)!
(2n−1)!
Hence absolutely convergent.
1 3
= lim n 3 = 1. Inconclusive.
35. (a) lim (n+1)
1
n→∞ n+1
n→∞
n3
n+1 = lim n+1 = 1 < 1. Absolutely convergent.
(b) lim 2n+1
n
2
n→∞ 2n
n→∞
2n
(−3)n √
p n
(c) lim (−3)n+1
= 3 > 1. Divergent.
n−1 = lim 3
n+1
n→∞
√
n→∞
n
√n+1 q
2
n+1 1+n2
√
(d) lim 1+(n+1)
=
lim
= 1. Inconclusive.
n
n 1+(n+1)2
n→∞
n→∞
2
1+n
36. For k = 1, lim
n→∞
(n!)2
n!
= lim n! = ∞.
n→∞
For k ≥ 2, ratio test will show this series is (absolutely) convergent:
((n+1)!)2 (k(n+1))! ((n + 1)!)2 (kn)!
lim (n!)2 = lim
2
n→∞
(kn + k)!
(kn)! n→∞ (n!)
1
n→∞
(kn + k)(kn + k − 1) · · · (kn + 1)
1
n+1
= lim
n→∞ k (kn + k − 1)(kn + k − 2) · · · (kn + 1)
1
,k = 2
4
=
0 ,k > 2
= lim (n + 1)2
1
41.
p
n
|an | = L < 1, there exists a number r such that L < r < 1. By the
n→∞
p
defintion of limit, we know there exists some N such that n > N ⇒ | n |an |−L| <
∞
p
P
r − L ⇒ n |an | < r ⇒ |an | < rn . Then, as
rn is a convergent series, by
(i) Since lim
comparison test, so is
∞
P
n=1
|an |.
n=1
(ii) By the
defintion of limit,
we know there exists some N such that n > N ⇒
p
p
n
n
|L − |an || < L − 1 ⇒ |an | > 1 ⇒ |an | > 1. Then, lim an can never be zero.
n→∞
∞
P
1
n
n=1
(iii) It’s inconclusive because of the following examples:
q
is absolutely convergent. (Note that lim n n12 = lim
−2 lim
e
x→∞
ln x
x
=e
1
x
x→∞ 1
−2 lim
=e
1
x→∞ x
−2 lim
= 1)
n→∞
∞
P
n=1
is divergent.
n=1
1
2/x
x→∞ x
(−1)n
n
∞
P
− x2 ln x
= lim e
x→∞
1
n2
=
is conditionally convergent.
42. (a) Ratio Test:
4
4n
(4(n + 1))!(1103 + 26390(n + 1))
(n!)
396
lim ·
n→∞
((n + 1)!)4 3964(n+1)
(4n)!(1103 + 26390n) (4n + 4)(4n + 3)(4n + 2)(4n + 1) 1103 + 26390(n + 1)
= lim
n→∞
(n + 1)4 3964
1103 + 26390n
1
4n + 4
4n + 3
4n + 2
4n + 1
1103 + 26390(n + 1)
=
lim
· lim
· lim
· lim
· lim
4
n→∞
n→∞
n→∞
n→∞
n→∞
396
n+1
n+1
n+1
n+1
1103 + 26390n
44
=
<1
3964
(b)
9801
1
√
2 2 1103
≈ 3.14159273.
9801
1
√
2 2 1103+ 4!·27493
4
396
2
≈ 3.1415926535897939. π ≈ 3.1415926535897932.