2
1
√
1− √ + 2
1
−2 n+1
n
n n
=− .
18. lim
= lim
2
1
1
1 − n − 3n
3
− −3
2
n
n
n2
24.
p
n 2 − (n 2 − 4n)
n 2 − 4n = lim
√
n + n 2 − 4n
4
4n
= lim
r
= 2.
= lim
√
4
n + n 2 − 4n
1+ 1−
n
lim n −
27.
(n!)2
(1 · 2 · 3 · · · n)(1 · 2 · 3 · · · n)
=
(2n)!
1 · 2 · 3 · · · n · (n + 1) · (n + 2) · · · 2n
n
1
2
3
n
1
=
.
·
·
···
≤
n+1 n+2 n+3
n+n
2
Thus lim an = 0.
an =
√
√
31. Let a1 = 3 and an+1 = 15 + 2an for n = 1, 2, 3, . . .. Then we have a2 = 21 > 3 = a1 .
If ak+1 > ak for some k, then
ak+2 =
p
15 + 2ak+1 >
p
15 + 2ak = ak+1 .
Thus, {an } is increasing by induction. Observe that
a1 < 5 and a2 < 5. If ak < 5 then
p
p
√
ak+1 = 15 + 2ak < 15 + 2(5) = 25 = 5.
Therefore, an < 5 for all n, by induction. Since {an } is increasing and bounded above, it
converges. Let lim an = a. Then
a=
√
15 + 2a ⇒ a 2 − 2a − 15 = 0 ⇒ a = −3, or a = 5.
Since a > a1 , we must have lim an = 5.
12. Let
∞
X
1
1
1
1
=
+
+
+ ···.
(2n − 1)(2n + 1)
1×3 3×5 5×7
n=1
1
1
1
1
Since
=
−
, the partial sum is
(2n − 1)(2n + 1)
2 2n − 1 2n + 1
1
1 11 1
sn =
−
1−
+
+···
2
3
2 3 5
1 1 1
1 1
1
−
−
+
+
2 2n − 3 2n − 1
2 2n − 1 2n + 1
1
1 =
1−
.
2
2n + 1
Hence,
∞
X
n=1
1
1
= lim sn = .
(2n − 1)(2n + 1)
2
P
n(n + 1)
2
, the given series is ∞
which converges
n=1
2
n(n + 1)
to 2 by the result of Example 3 of this section.
20. Since 1 + 2 + 3 + · · · + n =
21. The total distance is
#
2
3
3
+···
2+2 2× +2×
4
4
"
#
2
3
3
3
=2+2×
1+ +
+···
2
4
4
"
=2 +
3
3
1−
4
= 14 metres.
2m
Fig. 2-21
P
1
an diverges and {bn } is bounded, then an bn diverges” is FALSE. Let an = and
n
P
P
P
1
1
. Then
an = ∞ and 0 ≤ bn ≤ 1/2. But
an bn =
which
bn =
n+1
n(n + 1)
converges by Example 3.
30. “If
P
P
P
31. “If an P
> 0 and an converges, then an2 converges” is TRUE.
Since an converges, therefore lim an = 0.
Thus there exists N such that 0 < an ≤ 1 for n ≥ N . Thus 0 < an2 ≤ an for n ≥ N .
n
n
X
X
2
If Sn =
ak and sn =
ak , then {Sn } is increasing and bounded above:
k=N
k=N
Sn ≤ sn ≤
Thus
∞
X
k=N
ak2
converges, and so
∞
X
k=1
∞
X
k=1
ak < ∞.
ak2 converges.
27.
P
P
(−1)n
is a couna) “ an converges implies (−1)n an converges” is FALSE. an =
n
terexample.
P
P
P
b) “ an converges and (−1)n an converges implies
an converges absolutely” is
FALSE. The series of Exercise 25 is a counterexample.
P
P
c) “ an converges absolutely implies (−1)n an converges absolutely” is TRUE, because
|(−1)n an | = |an |.
41. Trying to apply the ratio test to
X 22n (n!)2
, we obtain
(2n)!
ρ = lim
(2n)!
4(n + 1)2
22n+2 ((n + 1)!)2
· 2n
=
lim
= 1.
(2n + 2)!
2 (n!)2
(2n + 2)(2n + 1)
Thus the ratio test provides no information. However,
22n (n!)2
[2n(2n − 2) · · · 6 · 4 · 2]2
=
(2n)!
2n(2n − 1)(2n − 2) · · · 3 · 2 · 1
2n − 2
4 2
2n
·
· · · · · · > 1.
=
2n − 1 2n − 3
3 1
Since the terms exceed 1, the series diverges to infinity.
n 2
∞ X
n
39.
converges by the root test of Exercise 31 since
n+1
n=1
σ = lim
n→∞
n
n+1
n 2 1/n
= lim n→∞
1
1+
=
1 n
n
1
< 1.
e
23. Apply the ratio test to
X (2x + 3)n
:
n 1/3 4n
1/3 4n x + 3 (2x + 3)n+1
n
|2x
+
3|
2
=
ρ = lim ·
=
.
(n + 1)1/3 4n+1 (2x + 3)n 4
2
3 7
1
The series converges absolutely if x + < 2, that is, if − < x < . By the alternating
2
2
2
7
series test it converges conditionally at x = − . It diverges elsewhere.
2
1
1 n
24. Let an =
1+
. Apply the ratio test
n
x
1 1 1 n+1 n 1 −n ρ = lim 1
+
1+
×
1+
=
<1
n+1
x
1
x
x
if and only if |x + 1| < |x|, that is, −2 <
∞
X
n=1
an =
∞
X
(−1)n
n=1
n
1
1
1
< 0 ⇒ x < − . If x = − , then
x
2
2
, which converges conditionally. Thus, the series converges absolutely
1
1
if x < − , converges conditionally if x = − and diverges elsewhere. It is undefined at
2
2
x = 0.
29. Applying the ratio test to
X
X (2n)!x n
=
an x n , we obtain
2n
2
2 (n!)
ρ = lim |x|
(2n + 2)(2n + 1)
= |x|.
4(n + 1)2
an x n converges absolutely if −1 < x < 1, and diverges if x > 1 or x < −1.
1
, so the given series definitely
In Exercise 36 of Section 9.3 it was shown that an ≥
2n
diverges at x = 1 and may at most converge conditionally at x = −1. To see whether it
does converge at −1, we write, as in Exercise 36 of Section 9.3,
Thus
P
1 × 2 × 3 × 4 × · · · × 2n
(2 × 4 × 6 × 8 × · · · × 2n)2
1 × 3 × 5 × · · · × (2n − 1)
=
2 × 4 × 6 × · · · × (2n − 2) × 2n
2n − 3 2n − 1
1 3
×
= × ×···×
2
4
2n
−
2
2n
1
1
1
1
1−
··· 1−
1−
.
= 1−
2
4
2n − 2
2n
(2n)!
an =
22n (n!)2
=
It is evident that an decreases as n increases. To see whether lim an = 0, take logarithms
and use the inequality ln(1 + x) ≤ x:
1
1
1
ln an = ln 1 −
+ ln 1 −
+ · · · + ln 1 −
2
4
2n
1 1
1
≤ − − −···−
2 4
2n 1
1
1
=−
1+ +···+
→ −∞ as n → ∞.
2
2
n
Thus lim an = 0, and the given series converges conditionally at x = −1 by the alternating
series test.