Solution of Exercise 11.2
EX.2
From the two figure, we can see that R6
1 f (x)dx <
5
P
i=1
ai &R6
1 f (x)dx >
6
P
i=2
ai
Thus,
6
P
i=2
ai <R6
1 f (x)dx <
5
P
i=1
ai.
EX.6
f (x) = √x+41 = (x + 4)−1/2 is continuous, positive, and decreasing on [1, ∞) Z ∞
1
(x+4)−1/2dx = lim
t→∞
Z t 1
(x+4)−1/2dx = lim
t→∞[2(x+4)1/2]t1 = lim
t→∞[2√
t + 4−2√
5] = ∞
So,
∞
P
n=1
√1
n+4 is divergent by the integral test.
EX.18
f (x) = x3x−42−2x = 2x +x−21 is continuous, positive, and decreasing on [3, ∞) Z ∞
3
3x − 4
x2 − 2xdx = lim
t→∞
Z t 3
[2 x+ 1
x − 2]dx = lim
t→∞[2 ln x+ln(x−2)]t3 = lim
t→∞[2 ln t+ln(t−2)−2 ln 3] = ∞ So,
∞
P
n=3 3n−4
n2−2n is divergent by the integral test.
EX.21
f (x) = x ln x1 is continuous, positive, and decreasing on [2, ∞) (f0(x) = −x1+ln x2(ln x)2 < 0 for x > 2)
1
Z ∞ 2
1
x ln xdx = lim
t→∞[ln(ln x)]t2 = lim
t→∞[ln(ln t) − ln(ln 2)] = ∞ So,
∞
P
n=2 1
n ln n is divergent by the integral test.
EX.22
f (x) = x(ln x)1 2 is continuous, positive, and decreasing on [2, ∞) Z ∞
2
1
x(ln x)2dx = lim
t→∞[−1
ln x]t2 = − lim
t→∞[ln t − ln 2] = 1 ln2 So,
∞
P
n=2 1
n(ln n)2 is convergent by the integral test.
EX.29
By EX.21, we know that when p = 1 the series
∞
P
n=2 1
n(ln n)p is divergent. So we may assume that p 6= 1.
f (x) = x(ln x)1 p is continuous, positive on [2, ∞), and f0(x) = −x2p+ln x(ln x)p+1 < 0 if x > e−p, so that f is eventually decreasing.
Z ∞ 2
1
x(ln x)pdx = lim
t→∞[(ln x)1−p
1 − p ]t2 = lim
t→∞[(ln t)1−p
1 − p − (ln 2)1−p 1 − p ] which exists whenever 1 − p < 0, so the series is convergent for p > 1.
EX.30
f (x) = x ln x[ln(ln x)]1 p is continuous, positive on [3, ∞). For p ≥ 0, it is clearly that f is decreasing on [3, ∞); and for p < 0, we can verify that f0(x) < 0 whenever x is greater than some l ∈ R. So we apply the integral test now.
I = Z ∞
3
1
x ln x[ln(ln x)]pdx = lim
t→∞
Z t 3
[ln(ln x)]−p
x ln x dx = lim
t→∞[[ln(ln x)]−p+1
−p + 1 ]t3 (for p 6= 1)
= lim
t→∞[[ln(ln t)]−p+1−p+1 = [ln(ln 3)]−p+1−p+1], which exists whenever −p + 1 > 0 ⇔ p > 1.
2
If p = 1, then I = lim
t→∞[ln(ln(ln x))]t3 = ∞.
So, the series is convergent for p > 1.
EX.32
If p ≤ 0, lim
n→∞
ln n
np = ∞, then the series is divergent.
So, we may assume that p > 0, f (x) = ln xxp is continuous, positive on [2, ∞), and f0(x) = −xp−1(1−p ln x)x2p < 0 if x > e1/p, so that f is eventually decreasing.
Z ∞ 1
ln x
xp dx = lim
t→∞[x1−p[(1 − p) ln x − 1]
(1 − p)2 ]t1 (for p 6= 1) = (1−p)1 2[ lim
t→∞t1−p[(1 − p) ln t − 1] + 1], which exists whenever 1 − p < 0 ⇔ p > 1. So the series is convergent for p > 1.
EX.33
ζ(x) =
∞
P
n=1 1
nx, which is p-series with p = x. In this case, the domain of ζ is the set of real numbers x such that the series is convergent. That is, ζ(x) is defined when x > 1.
EX.41
∞
P
n=1
n−1.001=
∞
P
n=1 1
n1.001 is a convergent p-series with p = 1.001 > 1. And Rn ≤
Z ∞ n
x−1.001dx = lim
t→∞[x−0.001
−0.001]tn= 1000 n0.001
And we want Rn < 5 × 10−9 ⇔ n10000.001 < 5 × 10−9 ⇔ n > (2 × 1011)1000 ≈ 1.07 × 1011,301.
EX.44
(a)
From the graph, Rn+1 1
1
xdx is less than the sum of the areas of the n rectan- gles.
3
So, we have
Z n+1 1
1
xdx = ln(n + 1) < 1 + 1
2+ · · · + 1 3 + 1
n
& ln n < ln(n + 1) < 1 +12+ · · · +13+1n, so 0 < 1 +12+ · · · +13+n1− ln n = tn.
(b)
It is clearly that Rn+1 n
1
xdx = ln(n + 1) − ln n > n+11 , so tn− tn+1 = [ln(n + 1) − ln n] − n+11 > 0, and this implies that {tn} is decreasing.
(c)
By (b) & (c), we know that {tn} is decreasing and tn > 0 for all n.
So, {tn} is a bounded monotonic sequence ⇒ {tn} is convergent.
EX.45
bln n = eln n·ln b = nln b = n− ln b1 . So,
∞
P
n=1
bln n =
∞
P
n=1 1
n− ln b is the p-series, which is convergent for all b such that − ln b > 1 ⇔ b < e−1 (with b > 0).
EX.46
Let sn=
n
P
i=1
(ci −i+11 ) = (c1 − 12) + (c2 − 13) + (c3 − 14) + · · · + (nc − n+11 )
= c1+c−12 +c−13 +c−14 + · · · +c−1n −n+11 = c + (c − 1)[12+13+14+ · · · +n1] −n+11 Thus,
∞
P
n=1
(nc −n+11 ) = lim
n→∞sn= lim
n→∞[c + (c − 1)
n
P
i=2 1
i −n+11 ],
the limit exists only if c − 1 = 0, so the series is convergent only if c = 1.
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