Calculus Solution of Section 11.1

Calculus Solution of Section 11.1
7. List the first five terms of the sequence. a1 = 3, an+1 = 2an − 1.
Sol. The sequence is {3, 5, 9, 17, 33, ...}.
13. Find a formula for the general term an of the sequence, assuming that the pattern of the first
few terms continues. {1, − 32 , 49 −
8
27 , ...}.
Sol. an = (− 23 )n−1 .
25. Determine whether the sequence { (2n−1)!
(2n+1)! } converges or diverges. If it converges, find the limit.
Sol. an =
(2n−1)!
(2n+1)!
=
(2n−1)!
(2n+1)(2n)(2n−1)!
=
1
(2n+1)(2n)
→ 0 as n → ∞. Converges.
1
35. Determine whether the sequence an = (1 + n2 ) n converges or diverges. If it converges, find the
limit.
1
Sol. an = (1 + n2 ) n ⇒ ln an =
1
n
ln(1 + n2 ). As n → ∞, ln an → 0. Thus, an → e0 = 1 as n → ∞.
Converges.
39. Determine whether the sequence an =
Sol. an =
n!
2n
=
1
2
·
2
2
·
3
2
· ··· ·
(n−1)
2
·
n
2
≥
1
2
n!
2n
·
n
2
converges or diverges. If it converges, find the limit.
[ for n > 1 ] = n4 → ∞ as n → ∞, so {an } diverges.
51. For what values of r is the sequence {nrn } convergent?
x
Sol. If |r| ≥ 1, then it is clear that {nrn } diverges. If |r| < 1 then lim xrx = lim −x =
x→∞
x→∞ r
1
rx
lim
= lim
= 0, so lim nrn = 0, and hence {nrn } converges whenever |r| < 1.
−x
x→∞ (− ln r)r
x→∞ − ln r
x→∞
59. Determine whether the sequence an = n2n+1 is increasing, decreasing, or not monotonic. Is the
sequence bounded?
Sol. an =
n
n2 +1
defines a decreasing sequence since for f (x) =
x
x2 +1
1 − x2
≤ 0 for x ≥ 1. The sequence is bounded since 0 < an ≤
(x2 + 1)2
q p
√ p √
√
61. Find the limit of the sequence { 2, 2 2, 2 2 2, ...}.
Sol. an = 2
2n −1
2n
, f 0 (x) =
1
2
(x2 + 1)(1) − x(2x)
=
(x2 + 1)2
for all n ≥ 1.
1
. lim an = lim 21− 2n = 2.
n→∞
n→∞
63. Show that the sequence defined by a1 = 1, an+1 = 3 −
1
an
is increasing and an < 3 for all n.
Deduce that {an } is convergent and find its limit.
Sol. We show by induction that {an } is increasing and bounded above by 3. Let Pn be the
proposition that an+1 > an and 0 < an < 3. Clearly P1 is true. Assume that Pn is true. Then
1
an+1 > an ⇒ − an+1
> − a1n . Now an+2 = 3 −
1
an+1
> 3−
1
an
= an+1 . Since an+1 = 3 −
1
an
and
0 < an < 3, we have 0 < an+1 < 3. Thus Pn+1 is ture. This proves that {an } is increasing and
bounded above by 3, and 1 = a1 < an < 3, that is, {an } is bounded, and hence convergent by the
Monotonic Sequence Theorem. If L = lim an , then lim an+1 = L, so L must satisfy L = 3 −
⇒ L2 − 3L + 1 = 0 ⇒ L =
√
3± 5
2 .
n→∞
But L > 1, so L =
1
n→∞
√
3+ 5
2 .
1
L

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