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January 31, 2005 12:53 L24-ch10 Sheet number 1 Page number 424 black
21. 2, (5/3)2, (6/4)3, (7/5)4, (8/6)5, . . . ; let y =[x+ 3x+ 1
]x, converges because
limx→+∞
ln y = limx→+∞
lnx+ 3x+ 11/x
= limx→+∞
2x2
(x+ 1)(x+ 3)= 2, so lim
n→+∞
[n+ 3n+ 1
]n= e2
22. −1, 0, (1/3)3, (2/4)4, (3/5)5, . . . ; let y = (1− 2/x)x, converges because
limx→+∞
ln y = limx→+∞
ln(1− 2/x)1/x
= limx→+∞
−21− 2/x
= −2, limn→+∞
(1− 2/n)n = limx→+∞
y = e−2
23.{2n− 12n
}+∞
n=1; limn→+∞
2n− 12n
= 1, converges
24.{n− 1n2
}+∞
n=1; limn→+∞
n− 1n2 = 0, converges
25.{(−1)n−1 1
3n
}+∞
n=1; limn→+∞
(−1)n−1
3n= 0, converges
26. {(−1)nn}+∞n=1; diverges because odd-numbered terms tend toward −∞,
even-numbered terms tend toward +∞.
27.{1n− 1
n+ 1
}+∞
n=1; limn→+∞
(1n− 1
n+ 1
)= 0, converges
28.{3/2n−1}+∞
n=1; limn→+∞
3/2n−1 = 0, converges
29.{√
n+ 1−√n+ 2
}+∞n=1; converges because
limn→+∞
(√n+ 1−
√n+ 2) = lim
n→+∞
(n+ 1)− (n+ 2)√n+ 1 +
√n+ 2
= limn→+∞
−1√n+ 1 +
√n+ 2
= 0
30.{(−1)n+1/3n+4}+∞
n=1; limn→+∞
(−1)n+1/3n+4 = 0, converges
31. an =
{+1 k even
−1 k oddoscillates; there is no limit point which attracts all of the an.
bn = cosn; the terms lie all over the interval [−1, 1] without any limit.
32. (a) No, because given N > 0, all values of f(x) are greater than N provided x is close enoughto zero. But certainly the terms 1/n will be arbitrarily close to zero, and when so thenf(1/n) > N , so f(1/n) cannot converge.
(b) f(x) = sin(π/x). Then f = 0 when x = 1/n and f �= 0 otherwise; indeed, the values of f arelocated all over the interval [−1, 1].
January 31, 2005 12:53 L24-ch10 Sheet number 3 Page number 426 black
426 Chapter 10
33. (a) 1, 2, 1, 4, 1, 6 (b) an =
{n, n odd
1/2n, n even(c) an =
{1/n, n odd
1/(n+ 1), n even
(d) In Part (a) the sequence diverges, since the even terms diverge to +∞ and the odd termsequal 1; in Part (b) the sequence diverges, since the odd terms diverge to +∞ and the eventerms tend to zero; in Part (c) lim
n→+∞an = 0.
34. The even terms are zero, so the odd terms must converge to zero, and this is true if and only if
limn→+∞
bn = 0, or −1 < b < 1.
35. limn→+∞
n√n = 1, so lim
n→+∞n√n3 = 13 = 1
37. limn→+∞
xn+1 =12
limn→+∞
(xn +
a
xn
)or L =
12
(L+
a
L
), 2L2 −L2 − a = 0, L =
√a (we reject −
√a
because xn > 0, thus L ≥ 0.)
38. (a) an+1 =√6 + an
(b) limn→+∞
an+1 = limn→+∞
√6 + an, L =
√6 + L, L2 − L− 6 = 0, (L− 3)(L+ 2) = 0,
L = −2 (reject, because the terms in the sequence are positive) or L = 3; limn→+∞
(d) Replace 0.5 in Part (a) with a0; then the sequence converges for −1 ≤ a0 ≤ 1, because ifa0 = ±1, then an = 1 for n ≥ 1; if a0 = 0 then an = 0 for n ≥ 1; and if 0 < |a0| < 1 thena1 = a2
0 > 0 and limn→+∞
an = limn→+∞
e2n−1 ln a1 = 0 since 0 < a1 < 1. This same argument
proves divergence to +∞ for |a| > 1 since then ln a1 > 0.
January 31, 2005 12:53 L24-ch10 Sheet number 4 Page number 427 black
Exercise Set 10.1 427
44. f(0.2) = 0.4, f(0.4) = 0.8, f(0.8) = 0.6, f(0.6) = 0.2 and then the cycle repeats, so the sequencedoes not converge.
45. (a) 30
00 5
(b) Let y = (2x + 3x)1/x, limx→+∞
ln y = limx→+∞
ln(2x + 3x)x
= limx→+∞
2x ln 2 + 3x ln 32x + 3x
= limx→+∞
(2/3)x ln 2 + ln 3(2/3)x + 1
= ln 3, so limn→+∞
(2n + 3n)1/n = eln 3 = 3
Alternate proof: 3 = (3n)1/n < (2n+3n)1/n < (2·3n)1/n = 3·21/n. Then apply the SqueezingTheorem.
46. Let f(x) = 1/(1 + x), 0 ≤ x ≤ 1. Take ∆xk = 1/n and x∗k = k/n then
an =n∑k=1
11 + (k/n)
(1/n) =n∑k=1
11 + x∗k
∆xk so limn→+∞
an =∫ 1
0
11 + x
dx = ln(1 + x)]1
0= ln 2
47. an =1
n− 1
∫ n
1
1xdx =
lnnn− 1
, limn→+∞
an = limn→+∞
lnnn− 1
= limn→+∞
1n= 0,(
apply L’Hopital’s Rule tolnnn− 1
), converges
48. (a) If n ≥ 1, then an+2 = an+1 + an, soan+2
an+1= 1 +
anan+1
.
(c) With L = limn→+∞
(an+2/an+1) = limn→+∞
(an+1/an), L = 1 + 1/L, L2 − L− 1 = 0,
L = (1±√5)/2, so L = (1 +
√5)/2 because the limit cannot be negative.
49.∣∣∣∣ 1n − 0
∣∣∣∣ = 1n< ε if n > 1/ε
(a) 1/ε = 1/0.5 = 2, N = 3 (b) 1/ε = 1/0.1 = 10, N = 11
16. f(x) = xe−2x, f ′(x) = (1− 2x)e−2x < 0 for x ≥ 1, so strictly decreasing.
17. f(x) =ln(x+ 2)x+ 2
, f ′(x) =1− ln(x+ 2)
(x+ 2)2< 0 for x ≥ 1, so strictly decreasing.
18. f(x) = tan−1 x, f ′(x) = 1/(1 + x2) > 0 for x ≥ 1, so strictly increasing.
January 31, 2005 12:53 L24-ch10 Sheet number 6 Page number 429 black
Exercise Set 10.2 429
19. f(x) = 2x2 − 7x, f ′(x) = 4x− 7 > 0 for x ≥ 2, so eventually strictly increasing.
20. f(x) = x3 − 4x2, f ′(x) = 3x2 − 8x = x(3x− 8) > 0 for x ≥ 3, so eventually strictly increasing.
21. f(x) =x
x2 + 10, f ′(x) =
10− x2
(x2 + 10)2< 0 for x ≥ 4, so eventually strictly decreasing.
22. f(x) = x+17x, f ′(x) =
x2 − 17x2 > 0 for x ≥ 5, so eventually strictly increasing.
23.an+1
an=
(n+ 1)!3n+1 · 3
n
n!=
n+ 13
> 1 for n ≥ 3, so eventually strictly increasing.
24. f(x) = x5e−x, f ′(x) = x4(5− x)e−x < 0 for x ≥ 6, so eventually strictly decreasing.
25. (a) Yes: a monotone sequence is increasing or decreasing; if it is increasing, then it is increas-ing and bounded above, so by Theorem 10.2.3 it converges; if decreasing, then use Theo-rem 10.2.4. The limit lies in the interval [1, 2].
(b) Such a sequence may converge, in which case, by the argument in Part (a), its limit is ≤ 2.But convergence may not happen: for example, the sequence {−n}+∞n=1 diverges.
26. The sequence {1} is monotone but not eventually strictly monotone. Let {xn}+∞n=1 be a sequencethat is monotone (let’s say increasing) but not eventually strictly monotone. Then there exists someN0 > 0 such that for all n ≥ N0 we have xn ≤ xn+1. Yet for any N > 0 it is not true that if n ≥ Nthen xn < xn+1. Since the sequence is monotone increasing we must have xn ≤ xn+1 for all n, yetinfinitely many times it is false that xn < xn+1. We conclude that for some n > N, xn = xn+1.Rephrasing, we can say that for any n > 0, there exists k, say kn, such that xkn = xkn+1. Thesequence has infinitely many pairs of repeated terms.
27. (a) an+1 =|x|n+1
(n+ 1)!=|x|
n+ 1|x|nn!
=|x|
n+ 1an
(b) an+1/an = |x|/(n+ 1) < 1 if n > |x| − 1.
(c) From Part (b) the sequence is eventually decreasing, and it is bounded below by 0, so byTheorem 10.2.4 it converges.
(d) If limn→+∞
an = L then from Part (a), L =|x|
limn→+∞
(n+ 1)L = 0.
(e) limn→+∞
|x|nn!
= limn→+∞
an = 0
28. (a)√2,
√2 +√2,
√2 +
√2 +√2
(b) a1 =√2 < 2 so a2 =
√2 + a1 <
√2 + 2 = 2, a3 =
√2 + a2 <
√2 + 2 = 2, and so on
indefinitely.
(c) a2n+1 − a2
n = (2 + an)− a2n = 2 + an − a2
n = (2− an)(1 + an)
(d) an > 0 and, from Part (b), an < 2 so 2 − an > 0 and 1 + an > 0 thus, from Part (c),
a2n+1 − a2
n > 0, an+1 − an > 0, an+1 > an; {an} is a strictly increasing sequence.
(e) The sequence is increasing and has 2 as an upper bound so it must converge to a limit L,lim
n→+∞an+1 = lim
n→+∞
√2 + an, L =
√2 + L, L2 − L − 2 = 0, (L − 2)(L + 1) = 0
thus limn→+∞
an = 2.
January 31, 2005 12:53 L24-ch10 Sheet number 7 Page number 430 black
430 Chapter 10
29. (a) If f(x) = 12 (x + 3/x), then f ′(x) = (x2 − 3)/(2x2) and f ′(x) = 0 for x =
√3; the minimum
value of f(x) for x > 0 is f(√3) =
√3. Thus f(x) ≥
√3 for x > 0 and hence an ≥
√3 for
n ≥ 2.
(b) an+1 − an = (3 − a2n)/(2an) ≤ 0 for n ≥ 2 since an ≥
√3 for n ≥ 2; {an} is eventually
decreasing.
(c)√3 is a lower bound for an so {an} converges; lim
n→+∞an+1 = lim
n→+∞12 (an + 3/an),
L = 12 (L+ 3/L), L2 − 3 = 0, L =
√3.
30. (a) The altitudes of the rectangles are ln k for k = 2 to n, and their bases all have length 1 sothe sum of their areas is ln 2 + ln 3 + · · · + lnn = ln(2 · 3 · · ·n) = lnn!. The area under the
curve y = lnx for x in the interval [1, n] is∫ n
1lnx dx, and
∫ n+1
1lnx dx is the area for x in
the interval [1, n+ 1] so, from the figure,∫ n
1lnx dx < lnn! <
∫ n+1
1lnx dx.
(b)∫ n
1lnx dx = (x lnx− x)
]n1= n lnn−n+1 and
∫ n+1
1lnx dx = (n+1) ln(n+1)−n so from
Part (a), n lnn − n + 1 < lnn! < (n + 1) ln(n + 1) − n, en lnn−n+1 < n! < e(n+1) ln(n+1)−n,
en lnne1−n < n! < e(n+1) ln(n+1)e−n,nn
en−1 < n! <(n+ 1)n+1
en
(c) From Part (b),[
nn
en−1
]1/n
<n√n! <
[(n+ 1)n+1
en
]1/n
,
n
e1−1/n <n√n! <
(n+ 1)1+1/n
e,
1e1−1/n <
n√n!n
<(1 + 1/n)(n+ 1)1/n
e,
but1
e1−1/n →1eand
(1 + 1/n)(n+ 1)1/n
e→ 1
eas n→ +∞ (why?), so lim
n→+∞
n√n!n
=1e.
31. n! >nn
en−1 ,n√n! >
n
e1−1/n , limn→+∞
n
e1−1/n = +∞ so limn→+∞
n√n! = +∞.
EXERCISE SET 10.3
1. (a) s1 = 2, s2 = 12/5, s3 =6225
, s4 =312125
sn =2− 2(1/5)n
1− 1/5=
52− 5
2(1/5)n,
limn→+∞
sn =52, converges
(b) s1 =14, s2 =
34, s3 =
74, s4 =
154
sn =(1/4)− (1/4)2n
1− 2= −1
4+
14(2n),
limn→+∞
sn = +∞, diverges
(c)1
(k + 1)(k + 2)=
1k + 1
− 1k + 2
, s1 =16, s2 =
14, s3 =
310
, s4 =13;
sn =12− 1
n+ 2, limn→+∞
sn =12, converges
January 31, 2005 12:53 L24-ch10 Sheet number 8 Page number 431 black
]�1= +∞, the series diverges by the Integral Test.
(b)∫ +∞
1
11 + 9x2 dx = lim
�→+∞
13tan−1 3x
]�1=
13(π/2− tan−1 3
),
the series converges by the Integral Test.
8. (a)∫ +∞
1
x
1 + x2 dx = lim�→+∞
12ln(1 + x2)
]�1= +∞, the series diverges by the Integral Test.
(b)∫ +∞
1(4 + 2x)−3/2dx = lim
�→+∞−1/√4 + 2x
]�1= 1/
√6,
the series converges by the Integral Test.
9.∞∑k=1
1k + 6
=∞∑k=7
1k, diverges because the harmonic series diverges.
10.∞∑k=1
35k
=∞∑k=1
35
(1k
), diverges because the harmonic series diverges.
11.∞∑k=1
1√k + 5
=∞∑k=6
1√k, diverges because the p-series with p = 1/2 ≤ 1 diverges.
January 31, 2005 12:53 L24-ch10 Sheet number 13 Page number 436 black
436 Chapter 10
12. limk→+∞
1e1/k = 1, the series diverges because lim
k→+∞uk = 1 �= 0.
13.∫ +∞
1(2x− 1)−1/3dx = lim
�→+∞
34(2x− 1)2/3
]�1= +∞, the series diverges by the Integral Test.
14.lnxx
is decreasing for x ≥ e, and∫ +∞
3
lnxx
= lim�→+∞
12(lnx)2
]�3= +∞,
so the series diverges by the Integral Test.
15. limk→+∞
k
ln(k + 1)= limk→+∞
11/(k + 1)
= +∞, the series diverges because limk→+∞
uk �= 0.
16.∫ +∞
1xe−x
2dx = lim
�→+∞−12e−x
2]�
1= e−1/2, the series converges by the Integral Test.
17. limk→+∞
(1 + 1/k)−k = 1/e �= 0, the series diverges.
18. limk→+∞
k2 + 1k2 + 3
= 1 �= 0, the series diverges.
19.∫ +∞
1
tan−1 x
1 + x2 dx = lim�→+∞
12(tan−1 x
)2]�
1= 3π2/32, the series converges by the Integral Test, since
d
dx
tan−1 x
1 + x2 =1− 2x tan−1 x
(1 + x2)2< 0 for x ≥ 1.
20.∫ +∞
1
1√x2 + 1
dx = lim�→+∞
sinh−1 x
]�1= +∞, the series diverges by the Integral Test.
21. limk→+∞
k2 sin2(1/k) = 1 �= 0, the series diverges.
22.∫ +∞
1x2e−x
3dx = lim
�→+∞−13e−x
3]�
1= e−1/3,
the series converges by the Integral Test (x2e−x3is decreasing for x ≥ 1).
23. 7∞∑k=5
k−1.01, p-series with p > 1, converges
24.∫ +∞
1sech2x dx = lim
�→+∞tanhx
]�1= 1− tanh(1), the series converges by the Integral Test.
25.1
x(lnx)pis decreasing for x ≥ ep, so use the Integral Test with
∫ +∞
ep
dx
x(lnx)pto get
lim�→+∞
ln(lnx)]�ep
= +∞ if p = 1, lim�→+∞
(lnx)1−p
1− p
]�ep
=
+∞ if p < 1
p1−p
p− 1if p > 1
Thus the series converges for p > 1.
January 31, 2005 12:53 L24-ch10 Sheet number 14 Page number 437 black
Exercise Set 10.4 437
26. If p > 0 set g(x) = x(lnx)[ln(lnx)]p, g′(x) = (ln(lnx))p−1 [(1 + lnx) ln(lnx) + p], and, for x > ee,
g′(x) > 0, thus 1/g(x) is decreasing for x > ee; use the Integral Test with∫ +∞
ee
dx
x(lnx)[ln(lnx)]p
to get
lim�→+∞
ln[ln(lnx)]]�ee
= +∞ if p = 1, lim�→+∞
[ln(lnx)]1−p
1− p
]�ee
=
+∞ if p < 1,
1p− 1
if p > 1
Thus the series converges for p > 1 and diverges for 0 < p ≤ 1. If p ≤ 0 then[ln(lnx)]p
x lnx≥ 1
x lnxfor x > ee so the series diverges.
27. Suppose Σ(uk + vk) converges; then so does Σ[(uk + vk) − uk], but Σ[(uk + vk) − uk] = Σvk, soΣvk converges which contradicts the assumption that Σvk diverges. Suppose Σ(uk−vk) converges;then so does Σ[uk − (uk − vk)] = Σvk which leads to the same contradiction as before.
28. Let uk = 2/k and vk = 1/k; then both Σ(uk + vk) and Σ(uk − vk) diverge; let uk = 1/k andvk = −1/k then Σ(uk + vk) converges; let uk = vk = 1/k then Σ(uk − vk) converges.
29. (a) diverges because∞∑k=1
(2/3)k−1 converges and∞∑k=1
1/k diverges.
(b) diverges because∞∑k=1
1/(3k + 2) diverges and∞∑k=1
1/k3/2 converges.
30. (a) converges because both∞∑k=2
1k(ln k)2
(Exercise 25) and∞∑k=2
1k2 converge.
(b) diverges, because+∞∑k=2
ke−k2converges (Integral Test), and, by Exercise 25,
+∞∑k=2
1k ln k
diverges
31. (a) 3∞∑k=1
1k2 −
∞∑k=1
1k4 = π2/2− π4/90 (b)
∞∑k=1
1k2 − 1− 1
22 = π2/6− 5/4
(c)∞∑k=2
1(k − 1)4
=∞∑k=1
1k4 = π4/90
32. (a) If S =∞∑k=1
uk and sn =n∑k=1
uk, then S − sn =∞∑
k=n+1
uk. Interpret uk, k = n+1, n+2, . . ., as
the areas of inscribed or circumscribed rectangles with height uk and base of length one forthe curve y = f(x) to obtain the result.
(b) Add sn =n∑k=1
uk to each term in the conclusion of Part (a) to get the desired result:
sn +∫ +∞
n+1f(x) dx <
+∞∑k=1
uk < sn +∫ +∞
n
f(x) dx
33. (a) In Exercise 32 above let f(x) =1x2 . Then
∫ +∞
n
f(x) dx = − 1x
]+∞
n
=1n;
use this result and the same result with n+ 1 replacing n to obtain the desired result.
January 31, 2005 12:53 L24-ch10 Sheet number 15 Page number 438 black
438 Chapter 10
(b) s3 = 1 + 1/4 + 1/9 = 49/36; 58/36 = s3 +14<
16π2 < s3 +
13= 61/36
(d) 1/11 <16π2 − s10 < 1/10
34. Apply Exercise 32 in each case:
(a) f(x) =1
(2x+ 1)2,
∫ +∞
n
f(x) dx =1
2(2n+ 1), so
146
<∞∑k=1
1(2k + 1)2
− s10 <142
(b) f(x) =1
k2 + 1,
∫ +∞
n
f(x) dx =π
2− tan−1(n), so
π/2− tan−1(11) <∞∑k=1
1k2 + 1
− s10 < π/2− tan−1(10)
(c) f(x) =x
ex,
∫ +∞
n
f(x) dx = (n+ 1)e−n, so 12e−11 <
∞∑k=1
k
ek− s10 < 11e−10
35. (a)∫ +∞
n
1x3 dx =
12n2 ; use Exercise 32(b) sw
(b)1
2n2 −1
2(n+ 1)2< 0.01 for n = 5.
(c) From Part (a) with n = 5 obtain 1.200 < S < 1.206, so S ≈ 1.203.
(2n+ 1)5 + 4(2n+ 1) ≥ 200, 2n+ 1 ≥ 3, n ≥ 1; n = 1, s1 = 0.20
45. (c) ak =1
2k − 1, an+1 =
12n+ 1
≤ 10−2, 2n+ 1 ≥ 100, n ≥ 49.5; n = 50
January 31, 2005 12:53 L24-ch10 Sheet number 23 Page number 446 black
446 Chapter 10
46. Suppose Σ|ak| converges, then limk→+∞
|ak| = 0 so |ak| < 1 for k ≥ K and thus |ak|2 < |ak|, a2k < |ak|
hence Σa2k converges by the Comparison Test.
47. (a)∑
(−1)k/k diverges, but∑
1/k2 converges;∑
(−1)k/√k converges, but
∑1/k diverges
(b) Let ak =(−1)kk
, then∑
a2k converges but
∑|ak| diverges,
∑ak converges.
48.∑
(1/kp) converges if p > 1 and diverges if p ≤ 1, so∞∑k=1
(−1)k 1kp
converges absolutely if p > 1,
and converges conditionally if 0 < p ≤ 1 since it satisfies the Alternating Series Test; it divergesfor p ≤ 0 since lim
k→+∞ak �= 0.
49. 1 +132 +
152 + · · ·=
[1 +
122 +
132 + · · ·
]−
[122 +
142 +
162 + · · ·
]
=π2
6− 1
22
[1 +
122 +
132 + · · ·
]=
π2
6− 1
4π2
6=
π2
8
50. Let A = 1− 122 +
132 −
142 + · · · ; since the series all converge absolutely,
π2
6−A = 2
122 + 2
142 + 2
162 + · · · = 1
2
(1 +
122 +
132 + · · ·
)=
12π2
6, so A =
12π2
6=
π2
12.
51. 1 +134 +
154 + · · ·=
[1 +
124 +
134 + · · ·
]−
[124 +
144 +
164 + · · ·
]
=π4
90− 1
24
[1 +
124 +
134 + · · ·
]=
π4
90− 1
16π4
90=
π4
96
52. Every positive integer can be written in exactly one of the three forms 2k − 1 or 4k − 2 or 4k,so a rearrangement is(1− 1
2− 1
4
)+
(13− 1
6− 1
8
)+
(15− 1
10− 1
12
)+ · · ·+
(1
2k − 1− 1
4k − 2− 1
4k
)+ · · ·
=(12− 1
4
)+
(16− 1
8
)+
(110− 1
12
)+ · · ·+
(1
4k − 2− 1
4k
)+ · · · = 1
2ln 2
53. (a) Write the series in the form(1− 1
2
)+
(23− 1
3
)+
(24− 1
4
)+ . . . =
+∞∑k=2
1k, which diverges.
(b) The alternating series test requires a sequence that i) alternates in sign, which this does, andii) decreases monotonely to 0, which this does not.
54. (a) 1.5
00 10
(b) Yes; since f(x) is decreasing for x ≥ 1
and limx→+∞
f(x) = 0, the series
satisfies the Alternating Series Test.
January 31, 2005 12:53 L24-ch10 Sheet number 24 Page number 447 black
Exercise Set 10.7 447
55. (a) The distance d from the starting point is
d = 180− 1802
+1803− · · · − 180
1000= 180
[1− 1
2+
13− · · · − 1
1000
].
From Theorem 10.6.2, 1− 12+
13− · · · − 1
1000differs from ln 2 by less than 1/1001 so
180(ln 2− 1/1001) < d < 180 ln 2, 124.58 < d < 124.77.
(b) The total distance traveled is s = 180 +1802
+1803
+ · · ·+ 1801000
, and from inequality (2) inSection 10.4,
∫ 1001
1
180x
dx < s < 180 +∫ 1000
1
180x
dx
180 ln 1001 < s < 180(1 + ln 1000)
1243 < s < 1424
EXERCISE SET 10.7
1. (a) f (k)(x) = (−1)ke−x, f (k)(0) = (−1)k; e−x ≈ 1− x+ x2/2 (quadratic), e−x ≈ 1− x (linear)
(b) f ′(x) = − sinx, f ′′(x) = − cosx, f(0) = 1, f ′(0) = 0, f ′′(0) = −1,cosx ≈ 1− x2/2 (quadratic), cosx ≈ 1 (linear)
(c) f ′(x) = cosx, f ′′(x) = − sinx, f(π/2) = 1, f ′(π/2) = 0, f ′′(π/2) = −1,sinx ≈ 1− (x− π/2)2/2 (quadratic), sinx ≈ 1 (linear)
(d) f(1) = 1, f ′(1) = 1/2, f ′′(1) = −1/4;√x = 1 +
12(x− 1)− 1
8(x− 1)2 (quadratic),
√x ≈ 1 +
12(x− 1) (linear)
2. (a) p2(x) = 1 + x+ x2/2, p1(x) = 1 + x
(b) p2(x) = 3 +16(x− 9)− 1
216(x− 9)2, p1(x) = 3 +
16(x− 9)
(c) p2(x) =π
3+√36
(x− 2)− 772
√3(x− 2)2, p1(x) =
π
3+√36
(x− 2)
(d) p2(x) = x, p1(x) = x
3. (a) f ′(x) =12x−1/2, f ′′(x) = −1
4x−3/2; f(1) = 1, f ′(1) =
12, f ′′(1) = −1
4;
√x ≈ 1 +
12(x− 1)− 1
8(x− 1)2
(b) x = 1.1, x0 = 1,√1.1 ≈ 1 +
12(0.1)− 1
8(0.1)2 = 1.04875, calculator value ≈ 1.0488088
4. (a) cosx ≈ 1− x2/2
(b) 2◦ = π/90 rad, cos 2◦ = cos(π/90) ≈ 1− π2
2 · 902 ≈ 0.99939077, calculator value ≈ 0.99939083
January 31, 2005 12:53 L24-ch10 Sheet number 25 Page number 448 black
448 Chapter 10
5. f(x) = tanx, 61◦ = π/3 + π/180 rad; x0 = π/3, f ′(x) = sec2 x, f ′′(x) = 2 sec2 x tanx;f(π/3) =
√3, f ′(π/3) = 4, f ′′(x) = 8
√3; tanx ≈
√3 + 4(x− π/3) + 4
√3(x− π/3)2,
tan 61◦ = tan(π/3 + π/180) ≈√3 + 4π/180 + 4
√3(π/180)2 ≈ 1.80397443,
calculator value ≈ 1.80404776
6. f(x) =√x, x0 = 36, f ′(x) =
12x−1/2, f ′′(x) = −1
4x−3/2;
f(36) = 6, f ′(36) =112
, f ′′(36) = − 1864
;√x ≈ 6 +
112
(x− 36)− 11728
(x− 36)2;
√36.03 ≈ 6 +
0.0312− (0.03)2
1728≈ 6.00249947917, calculator value ≈ 6.00249947938
7. f (k)(x) = (−1)ke−x, f (k)(0) = (−1)k; p0(x) = 1, p1(x) = 1− x, p2(x) = 1− x+12x2,
24. f(e) = 1, for k ≥ 1, f (k)(x) =(−1)k−1(k − 1)!
xk; f (k)(e) =
(−1)k−1(k − 1)!ek
;
p0(x) = 1, p1(x) = 1 +1e(x− e); p2(x) = 1 +
1e(x− e)− 1
2e2 (x− e)2;
p3(x) = 1 +1e(x− e)− 1
2e2 (x− e)2 +13e3 (x− e)3,
p4(x) = 1 +1e(x− e)− 1
2e2 (x− e)2 +13e3 (x− e)3 − 1
4e4 (x− e)4; 1 +n∑k=1
(−1)k−1
kek(x− e)k
25. (a) f(0) = 1, f ′(0) = 2, f ′′(0) = −2, f ′′′(0) = 6, the third MacLaurin polynomial for f(x) is f(x).(b) f(1) = 1, f ′(1) = 2, f ′′(1) = −2, f ′′′(1) = 6, the third Taylor polynomial for f(x) is f(x).
26. (a) f (k)(0) = k!ck for k ≤ n; the nth Maclaurin polynomial for f(x) is f(x).(b) f (k)(x0) = k!ck for k ≤ n; the nth Taylor polynomial about x = 1 for f(x) is f(x).
27. f (k)(0) = (−2)k; p0(x) = 1, p1(x) = 1− 2x,
p2(x) = 1− 2x+ 2x2, p3(x) = 1− 2x+ 2x2 − 43x3
4
-1
–0.6 0.6
28. f (k)(π/2) = 0 if k is odd, f (k)(π/2) is alternately 1
and −1 if k is even; p0(x) = 1, p2(x) = 1− 12(x− π/2)2,
p4(x) = 1− 12(x− π/2)2 +
124
(x− π/2)4,
p6(x) = 1− 12(x− π/2)2 +
124
(x− π/2)4 − 1720
(x− π/2)6
1.25
–1.25
^ i
January 31, 2005 12:53 L24-ch10 Sheet number 28 Page number 451 black
Exercise Set 10.7 451
29. f (k)(π) = 0 if k is odd, f (k)(π) is alternately −1
and 1 if k is even; p0(x) = −1, p2(x) = −1 +12(x− π)2,
p4(x) = −1 +12(x− π)2 − 1
24(x− π)4,
p6(x) = −1 +12(x− π)2 − 1
24(x− π)4 +
1720
(x− π)6
1.25
–1.25
0 o
30. f(0) = 0; for k ≥ 1, f (k)(x) =(−1)k−1(k − 1)!
(x+ 1)k,
f (k)(0) = (−1)k−1(k − 1)!; p0(x) = 0, p1(x) = x,
p2(x) = x− 12x2, p3(x) = x− 1
2x2 +
13x3
1.5
–1.5
–1 1
31. f (k)(x) = ex, |f (k)(x)| ≤ e1/2 < 2 on [0, 1/2], let M = 2,
e1/2 = 1 +12+
18+
148
+1
24 · 16 + · · ·+ 1n!2n
+Rn(1/2);
|Rn(1/2)| ≤M
(n+ 1)!(1/2)n+1 ≤ 2
(n+ 1)!(1/2)n+1 ≤ 0.00005 for n = 5;
e1/2 ≈ 1 +12+
18+
148
+1
24 · 16 +1
120 · 32 ≈ 1.64870, calculator value 1.64872
32. f(x) = ex, f (k)(x) = ex, |f (k)(x)| ≤ 1 on [−1, 0], |Rn(x)| ≤1
(n+ 1)!(1)n+1 =
1(n+ 1)!
< 0.5×10−3
if n = 6, so e−1 ≈ 1− 1 +12!− 1
3!+
14!− 1
5!+
16!≈ 0.3681, calculator value 0.3679
33. f (k)(ln 4) = 15/8 for k even, f (k)(ln 4) = 17/8 for k odd, which can be written as
f (k)(ln 4) =16− (−1)k
8;
n∑k=0
16− (−1)k8k!
(x− ln 4)k
34. (a) cosα ≈ 1− α2/2; x = r − r cosα = r(1− cosα) ≈ rα2/2
(b) In Figure Ex-36 let r = 4000 mi and α = 1/80 so that the arc has length 2rα = 100 mi.
Then x ≈ rα2/2 =40002 · 802 = 5/16 mi.
35. p(0) = 1, p(x) has slope −1 at x = 0, and p(x) is concave up at x = 0, eliminating I, II and IIIrespectively and leaving IV.
(c) |esin x − (1 + x)| < 0.01for − 0.14 < x < 0.14
0.01
0–0.15 0.15
(d) |esin x − (1 + x+ x2/2)| < 0.01for − 0.50 < x < 0.50
0.015
0–0.6 0.6
38. (a) f (k)(x) = ex ≤ eb,
|R2(x)| ≤ebb3
3!< 0.0005,
ebb3 < 0.003 if b ≤ 0.137 (by trial and
error with a hand calculator), so [0, 0.137].
(b)
0.20
0.002
0
39. (a) sinx = x− x3
3!+ 0 · x4 +R4(x),
|R4(x)| ≤|x|55!
< 0.5× 10−3 if |x|5 < 0.06,
|x| < (0.06)1/5 ≈ 0.569, (−0.569, 0.569)
(b) 0.0005
–0.0005
–0.57 0.57
January 31, 2005 12:53 L24-ch10 Sheet number 30 Page number 453 black
Exercise Set 10.8 453
40. M = 1, cosx = 1− x2
2!+
x4
4!+R5(x),
R5(x) ≤ 16! |x|6 ≤ 0.0005 if |x| < 0.8434
0
–0.0005
–0.85 0.85
41. f (5)(x) = − 3840(1 + x2)7
+3840x3
(1 + x2)5− 720x
(1 + x2)4, let M = 8400,
R4(x) ≤84004!|x|4 < 0.0005 if x < 0.0677
0
–0.00000005
–0.07 0.07
42. f(x) = ln(1 + x), f (4)(x) = −6/(1 + x)4, first assume |x| < 0.5, then we can calculate
M = 6/2−4 = 96, and |f(x)− p(x)| ≤ 64!|x|4 < 0.0005 if |x| < 0.2114
0
–0.0005
–0.22 0.22
EXERCISE SET 10.8
1. f (k)(x) = (−1)ke−x, f (k)(0) = (−1)k;∞∑k=0
(−1)kk!
xk
2. f (k)(x) = akeax, f (k)(0) = ak;∞∑k=0
ak
k!xk
3. f (k)(0) = 0 if k is odd, f (k)(0) is alternately πk and −πk if k is even;∞∑k=0
(−1)kπ2k
(2k)!x2k
January 31, 2005 12:53 L24-ch10 Sheet number 31 Page number 454 black
454 Chapter 10
4. f (k)(0) = 0 if k is even, f (k)(0) is alternately πk and −πk if k is odd;∞∑k=0
(−1)kπ2k+1
(2k + 1)!x2k+1
5. f (0)(0) = 0; for k ≥ 1, f (k)(x) =(−1)k+1(k − 1)!
(1 + x)k, f (k)(0) = (−1)k+1(k − 1)!;
∞∑k=1
(−1)k+1
kxk
6. f (k)(x) = (−1)k k!(1 + x)k+1 ; f
(k)(0) = (−1)kk!;∞∑k=0
(−1)kxk
7. f (k)(0) = 0 if k is odd, f (k)(0) = 1 if k is even;∞∑k=0
1(2k)!
x2k
8. f (k)(0) = 0 if k is even, f (k)(0) = 1 if k is odd;∞∑k=0
1(2k + 1)!
x2k+1
9. f (k)(x) =
{(−1)k/2(x sinx− k cosx) k even
(−1)(k−1)/2(x cosx+ k sinx) k odd, f (k)(0) =
{(−1)1+k/2k k even
0 k odd
∞∑k=0
(−1)k(2k + 1)!
x2k+2
10. f (k)(x) = (k + x)ex, f (k)(0) = k;∞∑k=1
1(k − 1)!
xk
11. f (k)(x0) = e;∞∑k=0
e
k!(x− 1)k
12. f (k)(x) = (−1)ke−x, f (k)(ln 2) = (−1)k 12;
∞∑k=0
(−1)k2 · k! (x− ln 2)k
13. f (k)(x) =(−1)kk!xk+1 , f (k)(−1) = −k!;
∞∑k=0
(−1)(x+ 1)k
14. f (k)(x) =(−1)kk!
(x+ 2)k+1 , f(k)(3) =
(−1)kk!5k+1 ;
∞∑k=0
(−1)k5k+1 (x− 3)k
15. f (k)(1/2) = 0 if k is odd, f (k)(1/2) is alternately πk and −πk if k is even;∞∑k=0
(−1)kπ2k
(2k)!(x− 1/2)2k
16. f (k)(π/2) = 0 if k is even, f (k)(π/2) is alternately −1 and 1 if k is odd;∞∑k=0
(−1)k+1
(2k + 1)!(x−π/2)2k+1
17. f(1) = 0, for k ≥ 1, f (k)(x) =(−1)k−1(k − 1)!
xk; f (k)(1) = (−1)k−1(k − 1)!;
∞∑k=1
(−1)k−1
k(x− 1)k
January 31, 2005 12:53 L24-ch10 Sheet number 32 Page number 455 black
Exercise Set 10.8 455
18. f(e) = 1, for k ≥ 1, f (k)(x) =(−1)k−1(k − 1)!
xk; f (k)(e) =
(−1)k−1(k − 1)!ek
;
1 +∞∑k=1
(−1)k−1
kek(x− e)k
19. geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ = |x|, so the interval of convergence is −1 < x < 1, converges
there to1
1 + x(the series diverges for x = ±1)
20. geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ = |x|2, so the interval of convergence is −1 < x < 1, converges
there to1
1− x2 (the series diverges for x = ±1)
21. geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ = |x−2|, so the interval of convergence is 1 < x < 3, converges
there to1
1− (x− 2)=
13− x
(the series diverges for x = 1, 3)
22. geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ = |x + 3|, so the interval of convergence is −4 < x < −2,
converges there to1
1 + (x+ 3)=
14 + x
(the series diverges for x = −4,−2)
23. (a) geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ = |x/2|, so the interval of convergence is −2 < x < 2,
converges there to1
1 + x/2=
22 + x
; (the series diverges for x = −2, 2)
(b) f(0) = 1; f(1) = 2/3
24. (a) geometric series, ρ = limk→+∞
∣∣∣∣uk+1
uk
∣∣∣∣ =∣∣∣∣x− 5
3
∣∣∣∣, so the interval of convergence is 2 < x < 8,
converges to1
1 + (x− 5)/3=
3x− 2
(the series diverges for x = 2, 8)
(b) f(3) = 3, f(6) = 3/4
25. ρ = limk→+∞
k + 1k + 2
|x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1,
∞∑k=0
(−1)kk + 1
converges by the Alternating Series Test; if x = 1,∞∑k=0
1k + 1
diverges. The radius of
convergence is 1, the interval of convergence is [−1, 1).
26. ρ = limk→+∞
3|x| = 3|x|, the series converges if 3|x| < 1 or |x| < 1/3 and diverges if |x| > 1/3. If
x = −1/3,∞∑k=0
(−1)k diverges, if x = 1/3,∞∑k=0
(1) diverges. The radius of convergence is 1/3, the
interval of convergence is (−1/3, 1/3).
January 31, 2005 12:53 L24-ch10 Sheet number 33 Page number 456 black
456 Chapter 10
27. ρ = limk→+∞
|x|k + 1
= 0, the radius of convergence is +∞, the interval is (−∞,+∞).
28. ρ = limk→+∞
k + 12|x| = +∞, the radius of convergence is 0, the series converges only if x = 0.
29. ρ = limk→+∞
5k2|x|(k + 1)2
= 5|x|, converges if |x| < 1/5 and diverges if |x| > 1/5. If x = −1/5,∞∑k=1
(−1)kk2
converges; if x = 1/5,∞∑k=1
1/k2 converges. Radius of convergence is 1/5, interval of convergence is
[−1/5, 1/5].
30. ρ = limk→+∞
ln kln(k + 1)
|x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1,
∞∑k=2
(−1)kln k
converges; if x = 1,∞∑k=2
1/(ln k) diverges (compare to∑
(1/k)). Radius of convergence
is 1, interval of convergence is [−1, 1).
31. ρ = limk→+∞
k|x|k + 2
= |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1,∞∑k=1
(−1)kk(k + 1)
converges;
if x = 1,∞∑k=1
1k(k + 1)
converges. Radius of convergence is 1, interval of convergence is [−1, 1].
32. ρ = limk→+∞
2k + 1k + 2
|x| = 2|x|, converges if |x| < 1/2, diverges if |x| > 1/2. If x = −1/2,∞∑k=0
−12(k + 1)
diverges; if x = 1/2,∞∑k=0
(−1)k2(k + 1)
converges. Radius of convergence is 1/2, interval of convergence
is (−1/2, 1/2].
33. ρ = limk→+∞
√k√
k + 1|x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1,
∞∑k=1
−1√kdiverges; if
x = 1,∞∑k=1
(−1)k−1√k
converges. Radius of convergence is 1, interval of convergence is (−1, 1].
34. ρ = limk→+∞
|x|2(2k + 2)(2k + 1)
= 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).
35. ρ = limk→+∞
|x|2(2k + 3)(2k + 2)
= 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).
36. ρ = limk→+∞
k3/2|x|3(k + 1)3/2
= |x|3, converges if |x| < 1, diverges if |x| > 1. If x = −1,∞∑k=0
1k3/2 converges;
if x = 1,∞∑k=0
(−1)kk3/2 converges. Radius of convergence is 1, interval of convergence is [−1, 1].
37. ρ = limk→+∞
3|x|k + 1
= 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).
January 31, 2005 12:53 L24-ch10 Sheet number 34 Page number 457 black
Exercise Set 10.8 457
38. ρ = limk→+∞
k(ln k)2|x|(k + 1)[ln(k + 1)]2
= |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, then, by
Exercise 10.4.25,∞∑k=2
−1k(ln k)2
converges; if x = 1,∞∑k=2
(−1)k+1
k(ln k)2converges. Radius of convergence
is 1, interval of convergence is [−1, 1].
39. ρ = limk→+∞
1 + k2
1 + (k + 1)2|x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1,
∞∑k=0
(−1)k1 + k2
converges; if x = 1,∞∑k=0
11 + k2 converges. Radius of convergence is 1, interval of convergence is
[−1, 1].
40. ρ = limk→+∞
12|x− 3| = 1
2|x− 3|, converges if |x− 3| < 2, diverges if |x− 3| > 2. If x = 1,
∞∑k=0
(−1)k
diverges; if x = 5,∞∑k=0
1 diverges. Radius of convergence is 2, interval of convergence is (1, 5).
41. ρ = limk→+∞
k|x+ 1|k + 1
= |x + 1|, converges if |x + 1| < 1, diverges if |x + 1| > 1. If x = −2,∞∑k=1
−1k
diverges; if x = 0,∞∑k=1
(−1)k+1
kconverges. Radius of convergence is 1, interval of convergence is
(−2, 0].
42. ρ = limk→+∞
(k + 1)2
(k + 2)2|x − 4| = |x − 4|, converges if |x − 4| < 1, diverges if |x − 4| > 1. If x = 3,
∞∑k=0
1/(k+1)2 converges; if x = 5,∞∑k=0
(−1)k/(k+1)2 converges. Radius of convergence is 1, interval
of convergence is [3, 5].
43. ρ = limk→+∞
(3/4)|x + 5| = 34|x + 5|, converges if |x + 5| < 4/3, diverges if |x + 5| > 4/3. If
x = −19/3,∞∑k=0
(−1)k diverges; if x = −11/3,∞∑k=0
1 diverges. Radius of convergence is 4/3, interval
of convergence is (−19/3,−11/3).
44. ρ = limk→+∞
(2k + 3)(2k + 2)k3
(k + 1)3|x− 2| = +∞, radius of convergence is 0,
series converges only at x = 2.
45. ρ = limk→+∞
k2 + 4(k + 1)2 + 4
|x+1|2 = |x+1|2, converges if |x+1| < 1, diverges if |x+1| > 1. If x = −2,
∞∑k=1
(−1)3k+1
k2 + 4converges; if x = 0,
∞∑k=1
(−1)kk2 + 4
converges. Radius of convergence is 1, interval of
convergence is [−2, 0].
January 31, 2005 12:53 L24-ch10 Sheet number 35 Page number 458 black
458 Chapter 10
46. ρ = limk→+∞
k ln(k + 1)(k + 1) ln k
|x − 3| = |x − 3|, converges if |x − 3| < 1, diverges if |x − 3| > 1. If
x = 2,∞∑k=1
(−1)k ln kk
converges; if x = 4,∞∑k=1
ln kk
diverges. Radius of convergence is 1, interval of
convergence is [2, 4).
47. ρ = limk→+∞
π|x− 1|2(2k + 3)(2k + 2)
= 0, radius of convergence +∞, interval of convergence (−∞,+∞).
48. ρ = limk→+∞
116|2x − 3| = 1
16|2x − 3|, converges if
116|2x − 3| < 1 or |x − 3/2| < 8, diverges if
|x− 3/2| > 8. If x = −13/2,∞∑k=0
(−1)k diverges; if x = 19/2,∞∑k=0
1 diverges. Radius of convergence
is 8, interval of convergence is (−13/2, 19/2).
49. ρ = limk→+∞
k√|uk| = lim
k→+∞
|x|ln k
= 0, the series converges absolutely for all x so the interval of
convergence is (−∞,+∞).
50. ρ = limk→+∞
2k + 1(2k)(2k − 1)
|x| = 0
so R = +∞.
51. If x ≥ 0, then cos√x = 1− (
√x)2
2!+
(√x)4
4!− (√x)6
6!+ · · · = 1− x
2!+
x2
4!− x3
6!+ · · ·; if x ≤ 0, then
cosh(√−x) = 1 +
(√−x)22!
+(√−x)44!
+(√−x)66!
+ · · · = 1− x
2!+
x2
4!− x3
6!+ · · · .
52. (a) 10
–1–1 1
53. By Exercise 76 of Section 3.6, the derivative of an odd (even) function is even (odd); hence allodd-numbered derivatives of an odd function are even, all even-numbered derivatives of an oddfunction are odd; a similar statement holds for an even function.
(a) If f(x) is an even function, then f (2k−1)(x) is an odd function, so f (2k−1)(0) = 0, and thusthe MacLaurin series coefficients a2k−1 = 0, k = 1, 2, · · ·.
(b) If f(x) is an odd function, then f (2k)(x) is an odd function, so f (2k)(0) = 0, and thus theMacLaurin series coefficients a2k = 0, k = 1, 2, · · ·.
54. By Theorem 10.4.3(b) both series converge or diverge together, so they have the same radius ofconvergence.
January 31, 2005 12:53 L24-ch10 Sheet number 36 Page number 459 black
Exercise Set 10.8 459
55. By Theorem 10.4.3(a) the series∑
(ck + dk)(x − x0)k converges if |x − x0| < R; if |x − x0| > Rthen
∑(ck+dk)(x−x0)k cannot converge, as otherwise
∑ck(x−x0)k would converge by the same
Theorem. Hence the radius of convergence of∑
(ck + dk)(x− x0)k is R.
56. Let r be the radius of convergence of∑
(ck + dk)(x − x0)k. If |x − x0| < min(R1, R2)then
∑ck(x − x0)k and
∑dk(x − x0)k converge, so
∑(ck + dk)(x − x0)k converges. Hence
r ≥ min(R1, R2) (to see that r > min(R1, R2) is possible consider the case ck = −dk = 1).If in addition R1 �= R2, and R1 < |x − x0| < R2 (or R2 < |x − x0| < R1) then∑
(ck + dk)(x − x0)k cannot converge, as otherwise all three series would converge. Thus inthis case r = min(R1, R2).
January 31, 2005 12:53 L24-ch10 Sheet number 45 Page number 468 black
468 Chapter 10
32.∫ 1/2
0(1 + x2)−1/4dx=
∫ 1/2
0
(1− 1
4x2 +
532
x4 − 15128
x6 + · · ·)dx
= x− 112
x3 +132
x5 − 15896
x7 + · · ·]1/2
0
= 1/2− 112
(1/2)3 +132
(1/2)5 − 15896
(1/2)7 + · · · ,
but15896
(1/2)7 < 0.5× 10−3 so∫ 1/2
0(1 + x2)−1/4dx ≈ 1/2− 1
12(1/2)3 +
132
(1/2)5 ≈ 0.4906
33. (a) Substitute x4 for x in the MacLaurin Series for ex to obtain+∞∑k=0
x4k
k!. The radius of conver-
gence is R = +∞.
(b) The first method is to multiply the MacLaurin Series for ex4by x3: x3ex
4=
+∞∑k=0
x4k+3
k!. The
second method involves differentiation:d
dxex
4= 4x3ex
4, so
x3ex4=
14d
dxex
4=
14d
dx
+∞∑k=0
x4k
k!=
14
+∞∑k=0
4kx4k−1
k!=
+∞∑k=0
x4k−1
(k − 1)!. Use the change of variable
j = k − 1 to show equality of the two series.
34. (a)x
(1− x)2= x
d
dx
[1
1− x
]= x
d
dx
[ ∞∑k=0
xk
]= x
[ ∞∑k=1
kxk−1
]=∞∑k=1
kxk
(b) − ln(1− x)=∫
11− x
dx− C =∫ [ ∞∑
k=0
xk
]dx− C
=∞∑k=0
xk+1
k + 1− C =
∞∑k=1
xk
k− C,− ln(1− 0) = 0 so C = 0.
(c) Replace x with −x in Part (b): ln(1 + x) = −+∞∑k=1
(−1)kk
xk =+∞∑k=1
(−1)k+1
kxk
(d)+∞∑k=1
(−1)k+1
kconverges by the Alternating Series Test.
(e) By Parts (c) and (d) and the remark,+∞∑k=1
(−1)k+1
kxk converges to ln(1 + x) for −1 < x ≤ 1.
35. (a) In Exercise 34(a), set x =13, S =
1/3(1− 1/3)2
=34
(b) In Part (b) set x = 1/4, S = ln(4/3)
36. (a) In Part (c) set x = 1, S = ln 2
(b) In Part (b) set x = (e− 1)/e, S = ln e = 1
37. (a) sinh−1 x=∫ (
1 + x2)−1/2dx− C =
∫ (1− 1
2x2 +
38x4 − 5
16x6 + · · ·
)dx− C
=(x− 1
6x3 +
340
x5 − 5112
x7 + · · ·)− C; sinh−1 0 = 0 so C = 0.
January 31, 2005 12:53 L24-ch10 Sheet number 46 Page number 469 black
Exercise Set 10.10 469
(b)(1 + x2)−1/2
= 1 +∞∑k=1
(−1/2)(−3/2)(−5/2) · · · (−1/2− k + 1)k!
(x2)k
= 1 +∞∑k=1
(−1)k 1 · 3 · 5 · · · (2k − 1)2kk!
x2k,
sinh−1 x = x+∞∑k=1
(−1)k 1 · 3 · 5 · · · (2k − 1)2kk!(2k + 1)
x2k+1
(c) R = 1
38. (a) sin−1 x=∫(1− x2)−1/2dx− C =
∫ (1 +
12x2 +
38x4 +
516
x6 + · · ·)dx− C
=(x+
16x3 +
340
x5 +5112
x7 + · · ·)− C, sin−1 0 = 0 so C = 0
(b)(1− x2)−1/2
= 1 +∞∑k=1
(−1/2)(−3/2)(−5/2) · · · (−1/2− k + 1)k!
(−x2)k
= 1 +∞∑k=1
(−1)k(1/2)k(1)(3)(5) · · · (2k − 1)k!
(−1)kx2k
= 1 +∞∑k=1
1 · 3 · 5 · · · (2k − 1)2kk!
x2k
sin−1 x= x+∞∑k=1
1 · 3 · 5 · · · (2k − 1)2kk!(2k + 1)
x2k+1
(c) R = 1
39. (a) y(t) = y0
∞∑k=0
(−1)k(0.000121)ktkk!
(b) y(1) ≈ y0(1− 0.000121t)]t=1
= 0.999879y0
(c) y0e−0.000121 ≈ 0.9998790073y0
40. (a) Ifct
m≈ 0 then e−ct/m ≈ 1− ct
m, and v(t) ≈
(1− ct
m
)(v0 +
mg
c
)− mg
c= v0 −
(cv0
m+ g
)t.
(b) The quadratic approximation is
v0 ≈(1− ct
m+
(ct)2
2m2
)(v0 +
mg
c
)− mg
c= v0 −
(cv0
m+ g
)t+
c2
2m2
(v0 +
mg
c
)t2.
41. θ0 = 5◦ = π/36 rad, k = sin(π/72)
(b) T ≈ 2π√
L
g= 2π
√1/9.8 ≈ 2.00709 (b) T ≈ 2π
√L
g
(1 +
k2
4
)≈ 2.008044621
(c) 2.008045644
January 31, 2005 12:53 L24-ch10 Sheet number 47 Page number 470 black
470 Chapter 10
42. The third order model gives the same result as the second, because there is no term of degree three
in (5). By the Wallis sine formula,∫ π/2
0sin4 φdφ =
1 · 32 · 4
π
2, and
T ≈ 4√
L
g
∫ π/2
0
(1 +
12k2 sin2 φ+
1 · 3222!
k4 sin4 φ
)dφ = 4
√L
g
(π
2+
k2
2π
4+
3k4
83π16
)
= 2π
√L
g
(1 +
k2
4+
9k4
64
)
43. (a) F =mgR2
(R+ h)2=
mg
(1 + h/R)2= mg
(1− 2h/R+ 3h2/R2 − 4h3/R3 + · · ·
)(b) If h = 0, then the binomial series converges to 1 and F = mg.
(c) Sum the series to the linear term, F ≈ mg − 2mgh/R.
(d)mg − 2mgh/R
mg= 1− 2h
R= 1− 2 · 29,028
4000 · 5280 ≈ 0.9973, so about 0.27% less.
44. (a) We can differentiate term-by-term:
y′ =∞∑k=1
(−1)kx2k−1
22k−1k!(k − 1)!=∞∑k=0
(−1)k+1x2k+1
22k+1(k + 1)!k!, y′′ =
∞∑k=0
(−1)k+1(2k + 1)x2k
22k+1(k + 1)!k!, and
xy′′ + y′ + xy =∞∑k=0
(−1)k+1(2k + 1)x2k+1
22k+1(k + 1)!k!+∞∑k=0
(−1)k+1x2k+1
22k+1(k + 1)!k!+∞∑k=0
(−1)kx2k+1
22k(k!)2,
xy′′ + y′ + xy =∞∑k=0
(−1)k+1x2k+1
22k(k!)2
[2k + 12(k + 1)
+1
2(k + 1)− 1
]= 0
(b) y′ =∞∑k=0
(−1)k(2k + 1)x2k
22k+1k!(k + 1)!, y′′ =
∞∑k=1
(−1)k(2k + 1)x2k−1
22k(k − 1)!(k + 1)!.
Since J1(x) =∞∑k=0
(−1)kx2k+1
22k+1k!(k + 1)!and x2J1(x) =
∞∑k=1
(−1)k−1x2k+1
22k−1(k − 1)!k!, it follows that
x2y′′ + xy′ + (x2 − 1)y
=∞∑k=1
(−1)k(2k + 1)x2k+1
22k(k − 1)!(k + 1)!+∞∑k=0
(−1)k(2k + 1)x2k+1
22k+1(k!)(k + 1)!+∞∑k=1
(−1)k−1x2k+1
22k−1(k − 1)!k!
−∞∑k=0
(−1)kx2k+1
22k+1k!(k + 1)!
=x
2− x
2+∞∑k=1
(−1)kx2k+1
22k−1(k − 1)!k!
(2k + 12(k + 1)
+2k + 1
4k(k + 1)− 1− 1
4k(k + 1)
)= 0.
(c) From Part (a), J ′0(x) =∞∑k=0
(−1)k+1x2k+1
22k+1(k + 1)!k!= −J1(x).
45. Suppose not, and suppose that k0 is the first integer for which ak �= bk. Thenak0x
k0 + ak0+1xk0+1 + . . . = bk0x
k0 + bk0+1xk0+1 + . . .. Divide by xk0 and let x→ 0 to show that
ak0 = bk0 which contradicts the assumption that they were not equal. Thus ak = bk for all k.
January 31, 2005 12:53 L24-ch10 Sheet number 48 Page number 471 black
Review Exercises, Chapter 10 471
REVIEW EXERCISES, CHAPTER 10
8. (a)∞∑k=0
f (k)(0)k!
xk (b)∞∑k=0
f (k)(x0)k!
(x− x0)k
9. (a) always true by Theorem 10.4.2(b) sometimes false, for example the harmonic series diverges but
∑(1/k2) converges
(c) sometimes false, for example f(x) = sinπx, ak = 0, L = 0(d) always true by the comments which follow Example 3(d) of Section 10.1
(e) sometimes false, for example an =12+ (−1)n 1
4(f) sometimes false, for example uk = 1/2
(g) always false by Theorem 10.4.3
(h) sometimes false, for example uk = 1/k, vk = 2/k
(i) always true by the Comparison Test
(j) always true by the Comparison Test
(k) sometimes false, for example∑
(−1)k/k(l) sometimes false, for example
∑(−1)k/k
10. (a) false, f(x) is not differentiable at x = 0, Definition 10.8.1
(b) true: sn = 1 if n is odd and s2n = 1 + 1/(n+ 1); limn→+∞
sn = 1
(c) false, lim ak �= 0
11. (a) an =n+ 2
(n+ 1)2 − n2 =n+ 2
((n+ 1) + n)((n+ 1)− n)=
n+ 22n+ 1
, limit = 1/2.
(b) an = (−1)n−1 n
2n+ 1, diverges by the Divergence Test (Theorem 10.4.1)
12. ak = √ak−1 = a1/2k−1 = a
1/4k−2 = · · · = a
1/2k−1
1 = c1/2k
(a) If c = 1/2 then limk→+∞
ak = 1. (b) if c = 3/2 then limk→+∞
ak = 1.
13. (a) an+1/an = (n + 1 − 10)4/(n − 10)4 = (n − 9)4/(n − 10)4. Since n − 9 > n − 10 for all n itfollows that (n− 9)4 > (n− 10)4 and thus that an+1/an > 1 for all n, hence the sequence isstrictly monotone increasing.
(b) If f(x) is a polynomial of degree n and k ≥ n then the Maclaurin polynomial of degree k isthe polynomial itself; if k < n then it is the truncated polynomial.