2. January 31, 2005 12:53 L24-ch10 Sheet number 2 Page number 425 black
Exercise Set 10.1 425
18. 0, −1, 0, 1, 0, . . . ; diverges
x2
19. e−1 , 4e−2 , 9e−3 , 16e−4 , 25e−5 , . . . ; lim x2 e−x = lim = 0, so lim n2 e−n = 0, converges
x→+∞ x→+∞ ex n→+∞
√ √ √ √
20. 1, 10 − 2, 18 − 3, 28 − 4,
40 − 5, . . . ;
3n 3 3
lim ( n2 + 3n − n) = lim √ = lim = , converges
n→+∞ n→+∞ n 2 + 3n + n n→+∞ 1 + 3/n + 1 2
x
x+3
21. 2, (5/3)2 , (6/4)3 , (7/5)4 , (8/6)5 , . . . ; let y = , converges because
x+1
x+3
ln 2x2 n+3
n
lim ln y = lim x + 1 = lim = 2, so lim = e2
x→+∞ x→+∞ 1/x x→+∞ (x + 1)(x + 3) n→+∞ n + 1
22. −1, 0, (1/3)3 , (2/4)4 , (3/5)5 , . . . ; let y = (1 − 2/x)x , converges because
ln(1 − 2/x) −2
lim ln y = lim = lim = −2, lim (1 − 2/n)n = lim y = e−2
x→+∞ x→+∞ 1/x x→+∞ 1 − 2/x n→+∞ x→+∞
+∞
2n − 1 2n − 1
23. ; lim = 1, converges
2n n=1
n→+∞ 2n
+∞
n−1 n−1
24. ; lim = 0, converges
n2 n=1
n→+∞ n2
+∞
1 (−1)n−1
25. (−1)n−1 ; lim = 0, converges
3n n=1
n→+∞ 3n
+∞
26. {(−1)n n}n=1 ; diverges because odd-numbered terms tend toward −∞,
even-numbered terms tend toward +∞.
+∞
1 1 1 1
27. − ; lim − = 0, converges
n n+1 n=1
n→+∞ n n+1
+∞
28. 3/2n−1 ; lim 3/2n−1
n=1 n→+∞
= 0, converges
√ √ +∞
29. n + 1 − n + 2 n=1 ; converges because
√ √ (n + 1) − (n + 2) −1
lim ( n + 1 − n + 2) = lim √ √ = lim √ √ =0
n→+∞ n→+∞ n+1+ n+2 n→+∞ n+1+ n+2
+∞
30. (−1)n+1 /3n+4 ; lim (−1)n+1 /3n+4
n=1 n→+∞
= 0, converges
+1 k even
31. an = oscillates; there is no limit point which attracts all of the an .
−1 k odd
bn = cos n; 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 enough
to zero. But certainly the terms 1/n will be arbitrarily close to zero, and when so then
f (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 are
located all over the interval [−1, 1].
3. January 31, 2005 12:53 L24-ch10 Sheet number 3 Page number 426 black
426 Chapter 10
n, n odd 1/n, n odd
33. (a) 1, 2, 1, 4, 1, 6 (b) an = (c) an =
n
1/2 , n even 1/(n + 1), n even
(d) In Part (a) the sequence diverges, since the even terms diverge to +∞ and the odd terms
equal 1; in Part (b) the sequence diverges, since the odd terms diverge to +∞ and the even
terms tend to zero; in Part (c) lim an = 0.
n→+∞
34. The even terms are zero, so the odd terms must converge to zero, and this is true if and only if
lim bn = 0, or −1 < b < 1.
n→+∞
√ √
n
35. lim n
n = 1, so lim n3 = 13 = 1
n→+∞ n→+∞
1 a 1 a √ √
37. lim xn+1 = lim xn + or L = L+ , 2L2 − L2 − a = 0, L = a (we reject − a
n→+∞ 2 n→+∞ xn 2 L
because xn > 0, thus L ≥ 0.)
√
38. (a) an+1 = 6 + an
√ √
(b) lim an+1 = lim 6 + an , L = 6 + L, L2 − L − 6 = 0, (L − 3)(L + 2) = 0,
n→+∞ n→+∞
L = −2 (reject, because the terms in the sequence are positive) or L = 3; lim an = 3.
n→+∞
1 2 1 2 3 1 2 3 4 3 2 5
39. (a) 1, + , + + , + + + = 1, , ,
4 4 9 9 9 16 16 16 16 4 3 8
1 1 1 1n+1
(c) an = (1 + 2 + · · · + n) = 2 n(n + 1) = , lim an = 1/2
n2 n 2 2 n n→+∞
1 4 1 4 9 1 4 9 16 5 14 15
40. (a) 1, + , + + , + + + = 1, , ,
8 8 27 27 27 64 64 64 64 8 27 32
1 2 1 1 1 (n + 1)(2n + 1)
(c) an = (1 + 22 + · · · + n2 ) = 3 n(n + 1)(2n + 1) = ,
n3 n 6 6 n2
1
lim an = lim (1 + 1/n)(2 + 1/n) = 1/3
n→+∞ n→+∞ 6
sin2 n 1
41. Let an = 0, bn = , cn = ; then an ≤ bn ≤ cn , lim an = lim cn = 0, so lim bn = 0.
n n n→+∞ n→+∞ n→+∞
n n n
1+n 3 n/2 + n
42. Let an = 0, bn = , cn = ; then (for n ≥ 2), an ≤ bn ≤ = cn ,
2n 4 2n
lim an = lim cn = 0, so lim bn = 0.
n→+∞ n→+∞ n→+∞
n
43. (a) a1 = (0.5)2 , a2 = a2 = (0.5)4 , . . . , an = (0.5)2
1
n
(c) lim an = lim e2 ln(0.5)
= 0, since ln(0.5) < 0.
n→+∞ n→+∞
(d) Replace 0.5 in Part (a) with a0 ; then the sequence converges for −1 ≤ a0 ≤ 1, because if
a0 = ±1, then an = 1 for n ≥ 1; if a0 = 0 then an = 0 for n ≥ 1; and if 0 < |a0 | < 1 then
n−1
a1 = a2 > 0 and lim an = lim e2
0
ln a1
= 0 since 0 < a1 < 1. This same argument
n→+∞ n→+∞
proves divergence to +∞ for |a| > 1 since then ln a1 > 0.
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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 sequence
does not converge.
45. (a) 30
0 5
0
ln(2x + 3x ) 2x ln 2 + 3x ln 3
(b) Let y = (2x + 3x )1/x , lim ln y = lim = lim
x→+∞ x→+∞ x x→+∞ 2x + 3x
x
(2/3) ln 2 + ln 3
= lim = ln 3, so lim (2n + 3n )1/n = eln 3 = 3
x→+∞ (2/3)x + 1 n→+∞
Alternate proof: 3 = (3n )1/n < (2n +3n )1/n < (2·3n )1/n = 3·21/n . Then apply the Squeezing
Theorem.
46. Let f (x) = 1/(1 + x), 0 ≤ x ≤ 1. Take ∆xk = 1/n and x∗ = k/n then
k
n n 1 1
1 1 1
an = (1/n) = ∆xk so lim an = dx = ln(1 + x) = ln 2
1 + (k/n) 1 + x∗
k
n→+∞ 0 1+x 0
k=1 k=1
n
1 1 ln n ln n 1
47. an = dx = , lim an = lim = lim = 0,
n−1 1 x n − 1 n→+∞ n→+∞ n − 1 n→+∞ n
ln n
apply L’Hˆpital’s Rule to
o , converges
n−1
an+2 an
48. (a) If n ≥ 1, then an+2 = an+1 + an , so =1+ .
an+1 an+1
(c) With L = lim (an+2 /an+1 ) = lim (an+1 /an ), L = 1 + 1/L, L2 − L − 1 = 0,
n→+∞ n→+∞
√ √
L = (1 ± 5)/2, so L = (1 + 5)/2 because the limit cannot be negative.
1 1
49. −0 = < if n > 1/
n n
(a) 1/ = 1/0.5 = 2, N = 3 (b) 1/ = 1/0.1 = 10, N = 11
(c) 1/ = 1/0.001 = 1000, N = 1001
n 1
50. −1 = < if n + 1 > 1/ , n > 1/ − 1
n+1 n+1
(a) 1/ − 1 = 1/0.25 − 1 = 3, N = 4 (b) 1/ − 1 = 1/0.1 − 1 = 9, N = 10
(c) 1/ − 1 = 1/0.001 − 1 = 999, N = 1000
1 1
51. (a) −0 = < if n > 1/ , choose any N > 1/ .
n n
n 1
(b) −1 = < if n > 1/ − 1, choose any N > 1/ − 1.
n+1 n+1
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428 Chapter 10
52. If |r| < 1 then lim rn = 0; if r > 1 then lim rn = +∞, if r < −1 then rn oscillates between
n→+∞ n→+∞
positive and negative values that grow in magnitude so lim rn does not exist for |r| > 1; if r = 1
n→+∞
then lim 1n = 1; if r = −1 then (−1)n oscillates between −1 and 1 so lim (−1)n does not exist.
n→+∞ n→+∞
EXERCISE SET 10.2
1 1 1
1. an+1 − an = − =− < 0 for n ≥ 1, so strictly decreasing.
n+1 n n(n + 1)
1 1 1
2. an+1 − an = (1 − ) − (1 − ) = > 0 for n ≥ 1, so strictly increasing.
n+1 n n(n + 1)
n+1 n 1
3. an+1 − an = − = > 0 for n ≥ 1, so strictly increasing.
2n + 3 2n + 1 (2n + 1)(2n + 3)
n+1 n 1
4. an+1 − an = − =− < 0 for n ≥ 1, so strictly decreasing.
4n + 3 4n − 1 (4n − 1)(4n + 3)
5. an+1 − an = (n + 1 − 2n+1 ) − (n − 2n ) = 1 − 2n < 0 for n ≥ 1, so strictly decreasing.
6. an+1 − an = [(n + 1) − (n + 1)2 ] − (n − n2 ) = −2n < 0 for n ≥ 1, so strictly decreasing.
an+1 (n + 1)/(2n + 3) (n + 1)(2n + 1) 2n2 + 3n + 1
7. = = = > 1 for n ≥ 1, so strictly increasing.
an n/(2n + 1) n(2n + 3) 2n2 + 3n
an+1 2n+1 1 + 2n 2 + 2n+1 1
8. = · = =1+ > 1 for n ≥ 1, so strictly increasing.
an 1 + 2n+1 2n 1 + 2n+1 1 + 2n+1
an+1 (n + 1)e−(n+1)
9. = = (1 + 1/n)e−1 < 1 for n ≥ 1, so strictly decreasing.
an ne−n
an+1 10n+1 (2n)! 10
10. = · = < 1 for n ≥ 1, so strictly decreasing.
an (2n + 2)! 10n (2n + 2)(2n + 1)
an+1 (n + 1)n+1 n! (n + 1)n
11. = · n = = (1 + 1/n)n > 1 for n ≥ 1, so strictly increasing.
an (n + 1)! n nn
2
an+1 5n+1 2n 5
12. = (n+1)2 · n = 2n+1 < 1 for n ≥ 1, so strictly decreasing.
an 2 5 2
13. f (x) = x/(2x + 1), f (x) = 1/(2x + 1)2 > 0 for x ≥ 1, so strictly increasing.
14. f (x) = 3 − 1/x, f (x) = 1/x2 > 0 for x ≥ 1, so strictly increasing.
1 + 1/x
15. f (x) = 1/(x + ln x), f (x) = − < 0 for x ≥ 1, so strictly decreasing.
(x + ln x)2
16. f (x) = xe−2x , f (x) = (1 − 2x)e−2x < 0 for x ≥ 1, so strictly decreasing.
ln(x + 2) 1 − ln(x + 2)
17. f (x) = , f (x) = < 0 for x ≥ 1, so strictly decreasing.
x+2 (x + 2)2
18. f (x) = tan−1 x, f (x) = 1/(1 + x2 ) > 0 for x ≥ 1, so strictly increasing.
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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.
x 10 − x2
21. f (x) = , f (x) = 2 < 0 for x ≥ 4, so eventually strictly decreasing.
x2 + 10 (x + 10)2
17 x2 − 17
22. f (x) = x + , f (x) = > 0 for x ≥ 5, so eventually strictly increasing.
x x2
an+1 (n + 1)! 3n n+1
23. = · = > 1 for n ≥ 3, so eventually strictly increasing.
an 3n+1 n! 3
24. f (x) = x5 e−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}+∞ diverges.
n=1
26. The sequence {1} is monotone but not eventually strictly monotone. Let {xn }+∞ be a sequence
n=1
that is monotone (let’s say increasing) but not eventually strictly monotone. Then there exists some
N0 > 0 such that for all n ≥ N0 we have xn ≤ xn+1 . Yet for any N > 0 it is not true that if n ≥ N
then xn < xn+1 . Since the sequence is monotone increasing we must have xn ≤ xn+1 for all n, yet
infinitely 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 . The
sequence has infinitely many pairs of repeated terms.
|x|n+1 |x| |x|n |x|
27. (a) an+1 = = = an
(n + 1)! n + 1 n! n+1
(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 by
Theorem 10.2.4 it converges.
|x|
(d) If lim an = L then from Part (a), L = L = 0.
n→+∞ lim (n + 1)
n→+∞
|x|n
(e) lim = lim an = 0
n→+∞ n! n→+∞
√ √ √
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) a2 − a2 = (2 + an ) − a2 = 2 + an − a2 = (2 − an )(1 + an )
n+1 n n n
(d) an > 0 and, from Part (b), an < 2 so 2 − an > 0 and 1 + an > 0 thus, from Part (c),
a2 − a2 > 0, an+1 − an > 0, an+1 > an ; {an } is a strictly increasing sequence.
n+1 n
(e) The sequence is increasing and has 2 as an upper bound so it must converge to a limit L,
√ √
lim an+1 = lim 2 + an , L = 2 + L, L2 − L − 2 = 0, (L − 2)(L + 1) = 0
n→+∞ n→+∞
thus lim an = 2.
n→+∞
7. January 31, 2005 12:53 L24-ch10 Sheet number 7 Page number 430 black
430 Chapter 10
√
29. (a) If f (x) = 1 (x + 3/x), then f (x) = (x2 − 3)/(2x2 ) and f (x) = 0 for x = 3; the minimum
2 √ √ √ √
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 − a2 )/(2an ) ≤ 0 for n ≥ 2 since an ≥ 3 for n ≥ 2; {an } is eventually
n
decreasing.
√
(c) 3 is a lower bound for an so {an } converges; lim an+1 = lim 1 (an + 3/an ),
n→+∞ 2
√ n→+∞
L = 2 (L + 3/L), L − 3 = 0, L = 3.
1 2
30. (a) The altitudes of the rectangles are ln k for k = 2 to n, and their bases all have length 1 so
the sum of their areas is ln 2 + ln 3 + · · · + ln n = ln(2 · 3 · · · n) = ln n!. The area under the
n n+1
curve y = ln x for x in the interval [1, n] is ln x dx, and ln x dx is the area for x in
1 1
n n+1
the interval [1, n + 1] so, from the figure, ln x dx < ln n! < ln x dx.
1 1
n n n+1
(b) ln x dx = (x ln x − x) = n ln n − n + 1 and ln x dx = (n + 1) ln(n + 1) − n so from
1 1 1
Part (a), n ln n − n + 1 < ln n! < (n + 1) ln(n + 1) − n, en ln n−n+1 < n! < e(n+1) ln(n+1)−n ,
nn (n + 1)n+1
en ln n e1−n < n! < e(n+1) ln(n+1) e−n , n−1
< n! <
e en
nn
1/n √
(n + 1)n+1
n
1/n
(c) From Part (b), < n! < ,
en−1 en
√
n √
n (n + 1)1+1/n 1 n
n! (1 + 1/n)(n + 1)1/n
< n! < , 1−1/n < < ,
e1−1/n e e n e
√
1 1 (1 + 1/n)(n + 1)1/n 1 n
n! 1
but → and → as n → +∞ (why?), so lim = .
e1−1/n e e e n→+∞ n e
nn √ n n n √
n
31. n! > n−1
, n! > 1−1/n , lim 1−1/n = +∞ so lim n! = +∞.
e e n→+∞ e n→+∞
EXERCISE SET 10.3
62 312 2 − 2(1/5)n 5 5
1. (a) s1 = 2, s2 = 12/5, s3 = , s4 = sn = = − (1/5)n ,
25 125 1 − 1/5 2 2
5
lim sn = , converges
n→+∞ 2
1 3 7 15 (1/4) − (1/4)2n 1 1
(b) s1 = , s2 = , s3 = , s4 = sn = = − + (2n ),
4 4 4 4 1−2 4 4
lim sn = +∞, diverges
n→+∞
1 1 1 1 1 3 1
(c) = − , s1 = , s2 = , s3 = , s4 = ;
(k + 1)(k + 2) k+1 k+2 6 4 10 3
1 1 1
sn = − , lim sn = , converges
2 n + 2 n→+∞ 2
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Exercise Set 10.3 431
2. (a) s1 = 1/4, s2 = 5/16, s3 = 21/64, s4 = 85/256
n−1 n
1 1 1 1 1 − (1/4)n 1 1 1
sn = 1+ + ··· + = = 1− ; lim sn =
4 4 4 4 1 − 1/4 3 4 n→+∞ 3
4n − 1
(b) s1 = 1, s2 = 5, s3 = 21, s4 = 85; sn = , diverges
3
(c) s1 = 1/20, s2 = 1/12, s3 = 3/28, s4 = 1/8;
n
1 1 1 1
sn = − = − , lim sn = 1/4
k+3 k+4 4 n + 4 n→+∞
k=1
1
3. geometric, a = 1, r = −3/4, sum = = 4/7
1 − (−3/4)
(2/3)3
4. geometric, a = (2/3)3 , r = 2/3, sum = = 8/9
1 − 2/3
7
5. geometric, a = 7, r = −1/6, sum = =6
1 + 1/6
6. geometric, r = −3/2, diverges
n
1 1 1 1
7. sn = − = − , lim sn = 1/3
k+2 k+3 3 n + 3 n→+∞
k=1
n
1 1 1 1
8. sn = − k+1 = − , lim sn = 1/2
2k 2 2 2n+1 n→+∞
k=1
n
1/3 1/3 1 1/3
9. sn = − = − , lim sn = 1/6
3k − 1 3k + 2 6 3n + 2 n→+∞
k=1
n+1 n+1 n+1
1/2 1/2 1 1 1
10. sn = − = −
k−1 k+1 2 k−1 k+1
k=2 k=2 k=2
n+1 n+3
1 1 1 1 1 1 1 3
= − = 1+ − − ; lim sn =
2 k−1 k−1 2 2 n+1 n+2 n→+∞ 4
k=2 k=4
∞ ∞
1
11. = 1/k, the harmonic series, so the series diverges.
k−2
k=3 k=1
(e/π)4 e4
12. geometric, a = (e/π)4 , r = e/π < 1, sum = = 3
1 − e/π π (π − e)
∞ ∞ k−1
4k+2 4 64
13. = 64 ; geometric, a = 64, r = 4/7, sum = = 448/3
7k−1 7 1 − 4/7
k=1 k=1
14. geometric, a = 125, r = 125/7, diverges
15. (a) Exercise 5 (b) Exercise 3 (c) Exercise 7 (d) Exercise 9
10. January 31, 2005 12:53 L24-ch10 Sheet number 10 Page number 433 black
Exercise Set 10.3 433
27. (a) Geometric series, a = x, r = −x2 . Converges for | − x2 | < 1, |x| < 1;
x x
S= = .
1 − (−x 2) 1 + x2
(b) Geometric series, a = 1/x2 , r = 2/x. Converges for |2/x| < 1, |x| > 2;
1/x2 1
S= = 2 .
1 − 2/x x − 2x
(c) Geometric series, a = e−x , r = e−x . Converges for |e−x | < 1, e−x < 1, ex > 1, x > 0;
e−x 1
S= −x
= x .
1−e e −1
1 1
28. Geometric series, a = sin x, r = − sin x. Converges for | − sin x| < 1, | sin x| < 2,
2 2
sin x 2 sin x
so converges for all values of x. S = = .
1 2 + sin x
1 + sin x
2
1 1 1 1 1 1 1 1 1 1 1 1 1
29. a2 = a1 + , a3 = a2 + = 2 a1 + 2 + , a4 = a3 + = 3 a1 + 3 + 2 + ,
2 2 2 2 2 2 2 2 2 2 2 2 2
1 1 1 1 1 1 1 1 1 1 1
a5 = a4 + = 4 a1 + 4 + 3 + 2 + , . . . , an = n−1 a1 + n−1 + n−2 + · · · + ,
2 2 2 2 2 2 2 2 2 2 2
∞ n
a1 1 1/2
lim an = lim + =0+ =1
n→+∞ n→+∞ 2n−1
n=1
2 1 − 1/2
√ √ √ √
k+1− k k+1− k 1 1
30. √ = √ √ =√ −√ ,
k 2+k k k+1 k k+1
n
1 1 1 1 1 1 1 1
sn = √ −√ = √ −√ + √ −√ + √ −√
k=1
k k+1 1 2 2 3 3 4
1 1 1
+··· + √ −√ =1− √ ; lim sn = 1
n n+1 n + 1 n→+∞
31. sn = (1 − 1/3) + (1/2 − 1/4) + (1/3 − 1/5) + (1/4 − 1/6) + · · · + [1/n − 1/(n + 2)]
= (1 + 1/2 + 1/3 + · · · + 1/n) − (1/3 + 1/4 + 1/5 + · · · + 1/(n + 2))
= 3/2 − 1/(n + 1) − 1/(n + 2), lim sn = 3/2
n→+∞
n n n n
1 1/2 1/2 1 1 1
32. sn = = − = −
k(k + 2) k k+2 2 k k+2
k=1 k=1 k=1 k=1
n n+2
1 1 1 1 1 1 1 3
= − = 1+ − − ; lim sn =
2 k k 2 2 n+1 n+2 n→+∞ 4
k=1 k=3
n n n n
1 1/2 1/2 1 1 1
33. sn = = − = −
(2k − 1)(2k + 1) 2k − 1 2k + 1 2 2k − 1 2k + 1
k=1 k=1 k=1 k=1
n n+1
1 1 1 1 1 1
= − = 1− ; lim sn =
2 2k − 1 2k − 1 2 2n + 1 n→+∞ 2
k=1 k=2
11. January 31, 2005 12:53 L24-ch10 Sheet number 11 Page number 434 black
434 Chapter 10
34. P0 P1 = a sin θ, a sin u cos u
P1
P1 P2 = a sin θ cos θ,
P2 P3 = a sin θ cos2 θ, P3 a sin u cos2 u
a sin u
P3 P4 = a sin θ cos3 θ, . . . u a sin u cos3 u
u
(see figure)
Each sum is a geometric series. u
P
P0 P2 P4
a
a sin θ
(a) P0 P1 + P1 P2 + P2 P3 + · · · = a sin θ + a sin θ cos θ + a sin θ cos2 θ + · · · =
1 − cos θ
(b) P0 P1 + P2 P3 + P4 P5 + · · · = a sin θ + a sin θ cos2 θ + a sin θ cos4 θ + · · ·
a sin θ a sin θ
= = = a csc θ
1 − cos2 θ sin2 θ
(c) P1 P2 + P3 P4 + P5 P6 + · · · = a sin θ cos θ + a sin θ cos3 θ + · · ·
a sin θ cos θ a sin θ cos θ
= = = a cot θ
1 − cos2 θ sin2 θ
θ θ θ θ θ/2
35. By inspection, − + − + ··· = = θ/3
2 4 8 16 1 − (−1/2)
1
36. A1 + A2 + A3 + · · · = 1 + 1/2 + 1/4 + · · · = =2
1 − (1/2)
2k A 2k B 2k 3k+1 − 2k+1 A + 2k 3k − 2k B
37. (b) + k+1 =
3k − 2 k 3 − 2k+1 (3k − 2k ) (3k+1 − 2k+1 )
3 · 6k − 2 · 22k A + 6k − 22k B (3A + B)6k − (2A + B)22k
= =
(3k − 2k ) (3k+1 − 2k+1 ) (3k − 2k ) (3k+1 − 2k+1 )
so 3A + B = 1 and 2A + B = 0, A = 1 and B = −2.
n n
2k 2k+1 2k
(c) sn = − k+1 = (ak − ak+1 ) where ak = .
3k − 2 k 3 − 2k+1 3k − 2 k
k=1 k=1
But sn = (a1 − a2 ) + (a2 − a3 ) + (a3 − a4 ) + · · · + (an − an+1 ) which is a telescoping sum,
2n+1 (2/3)n+1
sn = a1 − an+1 = 2 − , lim sn = lim 2 − = 2.
3n+1 − 2n+1 n→+∞ n→+∞ 1 − (2/3)n+1
∞
1 1 1
38. (a) geometric; 18/5 (b) geometric; diverges (c) − = 1/2
2 2k − 1 2k + 1
k=1
EXERCISE SET 10.4
∞ ∞ ∞
1 1/2 1 1/4 1 1
1. (a) = = 1; = = 1/3; + k = 1 + 1/3 = 4/3
2k 1 − 1/2 4k 1 − 1/4 2k 4
k=1 k=1 k=1
∞ ∞
1 1/5 1
(b) = = 1/4; =1 (Example 5, Section 10 .4);
5 k 1 − 1/5 k(k + 1)
k=1 k=1
∞
1 1
− = 1/4 − 1 = −3/4
5k k(k + 1)
k=1
12. January 31, 2005 12:53 L24-ch10 Sheet number 12 Page number 435 black
Exercise Set 10.4 435
∞ ∞
1 7 7/10
2. (a) = 3/4 (Exercise 10, Section 10.4); = = 7/9;
k2 − 1 10k−1 1 − 1/10
k=2 k=2
∞
1 7
so − = 3/4 − 7/9 = −1/36
k 2 − 1 10k−1
k=2
∞
9/7
(b) with a = 9/7, r = 3/7, geometric, 7−k 3k+1 = = 9/4;
1 − (3/7)
k=1
∞
2k+1 4/5
with a = 4/5, r = 2/5, geometric, = = 4/3;
5k 1 − (2/5)
k=1
∞ k+1
2
7−k 3k+1 − = 9/4 − 4/3 = 11/12
5k
k=1
3. (a) p = 3, converges (b) p = 1/2, diverges (c) p = 1, diverges (d) p = 2/3, diverges
4. (a) p = 4/3, converges (b) p = 1/4, diverges (c) p = 5/3, converges (d) p = π, converges
k
k2 + k + 3 1 1
5. (a) lim 2+1
= ; the series diverges. (b) lim 1+ = e; the series diverges.
k→+∞ 2k 2 k→+∞ k
1
(c) lim cos kπ does not exist; (d) lim = 0; no information
k→+∞ k→+∞ k!
the series diverges.
k
6. (a) lim = 0; no information (b) lim ln k = +∞; the series diverges.
k→+∞ ek k→+∞
√
1 k
(c) lim √ = 0; no information (d) lim √ = 1; the series diverges.
k→+∞ k k→+∞ k+3
+∞
1 1
7. (a) = lim ln(5x + 2) = +∞, the series diverges by the Integral Test.
1 5x + 2 →+∞ 5 1
+∞
1 1 1
(b) 2
dx = lim tan−1 3x = π/2 − tan−1 3 ,
1 1 + 9x →+∞ 3 1 3
the series converges by the Integral Test.
+∞
x 1
8. (a) 2
dx = lim ln(1 + x2 ) = +∞, the series diverges by the Integral Test.
1 1+x →+∞ 2 1
+∞ √ √
(b) (4 + 2x)−3/2 dx = lim −1/ 4 + 2x = 1/ 6,
1 →+∞ 1
the series converges by the Integral Test.
∞ ∞
1 1
9. = , diverges because the harmonic series diverges.
k+6 k
k=1 k=7
∞ ∞
3 3 1
10. = , diverges because the harmonic series diverges.
5k 5 k
k=1 k=1
∞ ∞
1 1
11. √ = √ , diverges because the p-series with p = 1/2 ≤ 1 diverges.
k=1
k + 5 k=6 k
13. January 31, 2005 12:53 L24-ch10 Sheet number 13 Page number 436 black
436 Chapter 10
1
12. lim = 1, the series diverges because lim uk = 1 = 0.
k→+∞ e1/k k→+∞
+∞
3
13. (2x − 1)−1/3 dx = lim (2x − 1)2/3 = +∞, the series diverges by the Integral Test.
1 →+∞ 4 1
+∞
ln x ln x 1
14. is decreasing for x ≥ e, and = lim (ln x)2 = +∞,
x 3 x →+∞ 2 3
so the series diverges by the Integral Test.
k 1
15. lim = lim = +∞, the series diverges because lim uk = 0.
k→+∞ ln(k + 1) k→+∞ 1/(k + 1) k→+∞
+∞
1
xe−x dx = lim − e−x = e−1 /2, the series converges by the Integral Test.
2 2
16.
1 →+∞ 2 1
17. lim (1 + 1/k)−k = 1/e = 0, the series diverges.
k→+∞
k2 + 1
18. lim = 1 = 0, the series diverges.
k→+∞ k 2 + 3
+∞
tan−1 x 1 2
19. 2
dx = lim tan−1 x = 3π 2 /32, the series converges by the Integral Test, since
1 1+x →+∞ 2 1
d tan−1 x 1 − 2x tan−1 x
= < 0 for x ≥ 1.
dx 1 + x2 (1 + x2 )2
+∞
1
20. √ dx = lim sinh−1 x = +∞, the series diverges by the Integral Test.
1 x2 +1 →+∞ 1
21. lim k 2 sin2 (1/k) = 1 = 0, the series diverges.
k→+∞
+∞
1
x2 e−x dx = lim − e−x = e−1 /3,
3 3
22.
1 →+∞ 3 1
the series converges by the Integral Test (x2 e−x is decreasing for x ≥ 1).
3
∞
23. 7 k −1.01 , p-series with p > 1, converges
k=5
+∞
24. sech2 x dx = lim tanh x = 1 − tanh(1), the series converges by the Integral Test.
1 →+∞ 1
+∞
1 dx
25. p
is decreasing for x ≥ ep , so use the Integral Test with to get
x(ln x) ep x(ln x)p
1−p
+∞
if p < 1
(ln x)
lim ln(ln x) = +∞ if p = 1, lim = 1−p
→+∞ ep →+∞ 1 − p ep p if p > 1
p−1
Thus the series converges for p > 1.
14. 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(ln x)[ln(ln x)]p , g (x) = (ln(ln x))p−1 [(1 + ln x) ln(ln x) + p], and, for x > ee ,
+∞
dx
g (x) > 0, thus 1/g(x) is decreasing for x > ee ; use the Integral Test with
ee x(ln x)[ln(ln x)]p
to get
1−p
+∞
if p < 1,
[ln(ln x)]
lim ln[ln(ln x)] = +∞ if p = 1, lim =
→+∞ ee →+∞ 1−p ee 1
if p > 1
p−1
[ln(ln x)]p 1
Thus the series converges for p > 1 and diverges for 0 < p ≤ 1. If p ≤ 0 then ≥
x ln x x ln x
for 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 , so
Σvk 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 and
vk = −1/k then Σ(uk + vk ) converges; let uk = vk = 1/k then Σ(uk − vk ) converges.
∞ ∞
29. (a) diverges because (2/3)k−1 converges and 1/k diverges.
k=1 k=1
∞ ∞
(b) diverges because 1/(3k + 2) diverges and 1/k 3/2 converges.
k=1 k=1
∞ ∞
1 1
30. (a) converges because both (Exercise 25) and converge.
k(ln k)2 k2
k=2 k=2
+∞ +∞
1
ke−k converges (Integral Test), and, by Exercise 25,
2
(b) diverges, because diverges
k ln k
k=2 k=2
∞ ∞ ∞
1 1 1 1
31. (a) 3 − = π 2 /2 − π 4 /90 (b) − 1 − 2 = π 2 /6 − 5/4
k2 k4 k 2 2
k=1 k=1 k=1
∞ ∞
1 1
(c) = = π 4 /90
(k − 1)4 k4
k=2 k=1
∞ n ∞
32. (a) If S = uk and sn = uk , then S − sn = uk . Interpret uk , k = n + 1, n + 2, . . ., as
k=1 k=1 k=n+1
the areas of inscribed or circumscribed rectangles with height uk and base of length one for
the curve y = f (x) to obtain the result.
n
(b) Add sn = uk to each term in the conclusion of Part (a) to get the desired result:
k=1
+∞ +∞ +∞
sn + f (x) dx < uk < sn + f (x) dx
n+1 k=1 n
+∞ +∞
1 1 1
33. (a) In Exercise 32 above let f (x) = . Then f (x) dx = − = ;
x2 n x n n
use this result and the same result with n + 1 replacing n to obtain the desired result.
15. January 31, 2005 12:53 L24-ch10 Sheet number 15 Page number 438 black
438 Chapter 10
1 1 1
(b) s3 = 1 + 1/4 + 1/9 = 49/36; 58/36 = s3 + < π 2 < s3 + = 61/36
4 6 3
1 2
(d) 1/11 < π − s10 < 1/10
6
34. Apply Exercise 32 in each case:
+∞ ∞
1 1 1 1 1
(a) f (x) = , f (x) dx = , so < − s10 <
(2x + 1)2 n 2(2n + 1) 46 (2k + 1)2 42
k=1
+∞
1 π
(b) f (x) = , f (x) dx = − tan−1 (n), so
k2 + 1 n 2
∞
1
π/2 − tan−1 (11) < − s10 < π/2 − tan−1 (10)
k2 +1
k=1
+∞ ∞
x k
(c) f (x) = , f (x) dx = (n + 1)e−n , so 12e−11 < − s10 < 11e−10
ex n ek
k=1
+∞
1 1
35. (a) dx = 2 ; use Exercise 32(b) sw
n x3 2n
1 1
(b) − < 0.01 for n = 5.
2n2 2(n + 1)2
(c) From Part (a) with n = 5 obtain 1.200 < S < 1.206, so S ≈ 1.203.
+∞
1 1 1 1
36. (a) dx = 3 ; choose n so that − < 0.005, n = 4; S ≈ 1.08
n x4 3n 3n3 3(n + 1)3
n n+1
1 1 1
37. (a) Let F (x) = , then dx = ln n and dx = ln(n + 1), u1 = 1 so
x 1 x 1 x
ln(n + 1) < sn < 1 + ln n.
(b) ln(1, 000, 001) < s1,000,000 < 1 + ln(1, 000, 000), 13 < s1,000,000 < 15
(c) s109 < 1 + ln 109 = 1 + 9 ln 10 < 22
(d) sn > ln(n + 1) ≥ 100, n ≥ e100 − 1 ≈ 2.688 × 1043 ; n = 2.69 × 1043
38. p-series with p = ln a; convergence for p > 1, a > e
39. x2 e−x is decreasing and positive for x > 2 so the Integral Test applies:
∞ ∞
x2 e−x dx = −(x2 + 2x + 2)e−x = 5e−1 so the series converges.
1 1
40. (a) f (x) = 1/(x3 + 1) is decreasing and continuous on the interval [1, +∞], so the Integral Test
applies.
(c) n 10 20 30 40 50
sn 0.681980 0.685314 0.685966 0.686199 0.686307
n 60 70 80 90 100
sn 0.686367 0.686403 0.686426 0.686442 0.686454
16. January 31, 2005 12:53 L24-ch10 Sheet number 16 Page number 439 black
Exercise Set 10.5 439
√ √
+∞
1 3 1 n3 + 1 3 2n − 1
(e) Set g(n) = dx = π + ln − tan−1 √ ; for n ≥ 13,
n x3 + 1 6 6 (n + 1)3 3 3
g(n) − g(n + 1) ≤ 0.0005; s13 + (g(13) + g(14))/2 ≈ 0.6865, so the sum ≈ 0.6865 to three
decimal places.
EXERCISE SET 10.5
∞
1 1 1 1
1. (a) ≤ 2 = 2, converges
5k 2 − k 5k − k 2 4k 4k 2
k=1
∞
3 3
(b) > , 3/k diverges
k − 1/4 k
k=1
∞ ∞
k+1 k 1 2 2 2
2. (a) 2−k
> 2 = , 1/k diverges (b) 4+k
< 4, converges
k k k k k k4
k=2 k=1
∞ ∞
1 1 1 5 sin2 k 5 5
3. (a) < k, converges (b) < , converges
3k + 5 3 3k k! k! k!
k=1 k=1
∞
ln k 1 1
4. (a) > for k ≥ 3, diverges
k k k
k=1
∞
k k 1 1
(b) > 3/2 = √ ; √ diverges
k 3/2 − 1/2 k k k=1
k
∞
4k 7 − 2k 6 + 6k 5
5. compare with the convergent series 1/k 5 , ρ = lim = 1/2, converges
k→+∞ 8k 7 + k − 8
k=1
∞
k
6. compare with the divergent series 1/k, ρ = lim = 1/9, diverges
k→+∞ 9k + 6
k=1
∞
3k
7. compare with the convergent series 5/3k , ρ = lim = 1, converges
k→+∞ 3k + 1
k=1
∞
k 2 (k + 3)
8. compare with the divergent series 1/k, ρ = lim = 1, diverges
k→+∞ (k + 1)(k + 2)(k + 5)
k=1
∞
1
9. compare with the divergent series ,
k=1
k 2/3
k 2/3 1
ρ = lim = lim = 1/2, diverges
k→+∞ (8k 2 − 3k) 1/3 k→+∞ (8 − 3/k)1/3
∞
10. compare with the convergent series 1/k 17 ,
k=1
k 17 1
ρ = lim 17
= lim = 1/217 , converges
k→+∞ (2k + 3) k→+∞ (2 + 3/k)17
17. January 31, 2005 12:53 L24-ch10 Sheet number 17 Page number 440 black
440 Chapter 10
3k+1 /(k + 1)! 3
11. ρ = lim = lim = 0, the series converges
k→+∞ 3k /k! k→+∞ k + 1
4k+1 /(k + 1)2 4k 2
12. ρ = lim = lim = 4, the series diverges
k→+∞ 4k /k 2 k→+∞ (k + 1)2
k
13. ρ = lim = 1, the result is inconclusive
k→+∞ k + 1
(k + 1)(1/2)k+1 k+1
14. ρ = lim = lim = 1/2, the series converges
k→+∞ k(1/2)k k→+∞ 2k
(k + 1)!/(k + 1)3 k3
15. ρ = lim = lim = +∞, the series diverges
k→+∞ k!/k 3 k→+∞ (k + 1)2
(k + 1)/[(k + 1)2 + 1] (k + 1)(k 2 + 1)
16. ρ = lim = lim = 1, the result is inconclusive.
k→+∞ k/(k 2 + 1) k→+∞ k(k 2 + 2k + 2)
3k + 2
17. ρ = lim = 3/2, the series diverges
k→+∞ 2k − 1
18. ρ = lim k/100 = +∞, the series diverges
k→+∞
k 1/k
19. ρ = lim = 1/5, the series converges
k→+∞ 5
20. ρ = lim (1 − e−k ) = 1, the result is inconclusive
k→+∞
21. Ratio Test, ρ = lim 7/(k + 1) = 0, converges
k→+∞
∞
22. Limit Comparison Test, compare with the divergent series 1/k
k=1
(k + 1)2
23. Ratio Test, ρ = lim = 1/5, converges
k→+∞ 5k 2
24. Ratio Test, ρ = lim (10/3)(k + 1) = +∞, diverges
k→+∞
25. Ratio Test, ρ = lim e−1 (k + 1)50 /k 50 = e−1 < 1, converges
k→+∞
∞
26. Limit Comparison Test, compare with the divergent series 1/k
k=1
∞
k3
27. Limit Comparison Test, compare with the convergent series 1/k 5/2 , ρ = lim = 1,
k→+∞ k 3 + 1
converges k=1
∞ ∞
4 4 4 4
28. kk
< k , kk
converges (Ratio Test) so converges by the Comparison Test
2+3 3 k 3 2 + k3k
k=1 k=1
![January 31, 2005 12:53 L24-ch10 Sheet number 2 Page number 425 black
Exercise Set 10.1 425
18. 0, −1, 0, 1, 0, . . . ; diverges
x2
19. e−1 , 4e−2 , 9e−3 , 16e−4 , 25e−5 , . . . ; lim x2 e−x = lim = 0, so lim n2 e−n = 0, converges
x→+∞ x→+∞ ex n→+∞
√ √ √ √
20. 1, 10 − 2, 18 − 3, 28 − 4,
40 − 5, . . . ;
3n 3 3
lim ( n2 + 3n − n) = lim √ = lim = , converges
n→+∞ n→+∞ n 2 + 3n + n n→+∞ 1 + 3/n + 1 2
x
x+3
21. 2, (5/3)2 , (6/4)3 , (7/5)4 , (8/6)5 , . . . ; let y = , converges because
x+1
x+3
ln 2x2 n+3
n
lim ln y = lim x + 1 = lim = 2, so lim = e2
x→+∞ x→+∞ 1/x x→+∞ (x + 1)(x + 3) n→+∞ n + 1
22. −1, 0, (1/3)3 , (2/4)4 , (3/5)5 , . . . ; let y = (1 − 2/x)x , converges because
ln(1 − 2/x) −2
lim ln y = lim = lim = −2, lim (1 − 2/n)n = lim y = e−2
x→+∞ x→+∞ 1/x x→+∞ 1 − 2/x n→+∞ x→+∞
+∞
2n − 1 2n − 1
23. ; lim = 1, converges
2n n=1
n→+∞ 2n
+∞
n−1 n−1
24. ; lim = 0, converges
n2 n=1
n→+∞ n2
+∞
1 (−1)n−1
25. (−1)n−1 ; lim = 0, converges
3n n=1
n→+∞ 3n
+∞
26. {(−1)n n}n=1 ; diverges because odd-numbered terms tend toward −∞,
even-numbered terms tend toward +∞.
+∞
1 1 1 1
27. − ; lim − = 0, converges
n n+1 n=1
n→+∞ n n+1
+∞
28. 3/2n−1 ; lim 3/2n−1
n=1 n→+∞
= 0, converges
√ √ +∞
29. n + 1 − n + 2 n=1 ; converges because
√ √ (n + 1) − (n + 2) −1
lim ( n + 1 − n + 2) = lim √ √ = lim √ √ =0
n→+∞ n→+∞ n+1+ n+2 n→+∞ n+1+ n+2
+∞
30. (−1)n+1 /3n+4 ; lim (−1)n+1 /3n+4
n=1 n→+∞
= 0, converges
+1 k even
31. an = oscillates; there is no limit point which attracts all of the an .
−1 k odd
bn = cos n; 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 enough
to zero. But certainly the terms 1/n will be arbitrarily close to zero, and when so then
f (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 are
located all over the interval [−1, 1].](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)