HOMEWORK #3 – SOLUTIONS
Page 646
10)
∞
3
∑ k ( ln k )
2
⇒ f ( x) =
k=2
3
x ( ln x )
2
> 0 and continuous.
1
2⎤
⎡
−3 ⎢ x ⋅ 2 ln x ⋅ + ( ln x ) ⎥
x
⎦=
f '( x) = ⎣
4
2
x ( ln x )
−3ln x [ 2 + ln x ] −3[ 2 + ln x ]
= 2
<0
4
3
x 2 ( ln x )
x ( ln x )
∴ f is decreasing ⇒ use Integral Test.
∞
t
3
∫ x ( ln x )
2
2
dx = lim ∫
t→∞
2
x ( ln x )
−3 ⎤
dx = lim
=
t→∞ ln x ⎥
⎦2
t
3
2
3 ⎤
3
⎡ −3
lim ⎢
+
=
⇒ series converges.
t→∞ ⎣ ln t
ln 2 ⎥⎦ ln 2
16)
ke− k
ke− k
k
k
∑ 4 + e− k ; 4 + e− k = 4ek2 + ek2 −k ≤ 4ek2
k=1
∞
k
Convergence of ∑ k 2 will prove convergence of
k=1 4e
the original series by the Comparison Test.
We will use the Integral Test to do that.
x
f ( x ) = x 2 > 0 and continuous for x ≥ 1.
4e
x2
x2
x2
⎡⎣1− 2x 2 ⎤⎦ 1− 2x 2
4e
4e − x ⋅ 8xe
f '( x) =
=
=
<0
2 x2
2 x2
x2
16e
16e
4e
∞
2
2
⇒ decreasing function.
∞
∞
1 − x2
1 − x2
⎡ −1 − x 2 ⎤
xe
dx
=
lim
xe
dx
=
lim
e ⎥ =
∫1 4
t→∞ ∫ 4
t→∞ ⎢
8
⎣
⎦1
1
−1 ⎡ 1 1 ⎤ 1
− =
⇒ convergence ⇒
8 ⎢⎣ e∞ e ⎥⎦ 8e
our original series converges.
56)
a)
c)
t
diverges
no conclusion
b)
d)
no conclusion
diverges
Page 654
4)
∞
k2
k2
=∞⇒
∑ ( −1) k + 1 ⇒ lim
k→∞ k + 1
k =1
original series diverges.
k +1
18)
4k
4 k 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 44
∑ ( −1) k! ; 0 < k! = 1⋅ 2 ⋅ 3⋅ 4 ⋅ 5k ⇒
k=3
4 ⋅ 4 ⋅ 4 ⋅ 4 128 128
boxed area < 1 and =
=
1⋅ 2 ⋅ 3⋅ 4
12
3k
k
4
128
≤
for k ≥ 4
k! 3k
128
4k
lim
= 0 ⇒ lim
=0
k→∞ 3k
k→∞ k!
ak+1
4 k+1 k!
4
=
⋅ k =
< 1, k ≥ 4
ak
k
+
1
!
4
k
+
1
( )
∞
∞
k
k
∞
4k
k 4
∑ ( −1) k! converges ⇒ ∑ ( −1) k! converges
k=4
k=3
k
22)
∞
1
1
; lim
=0
ln k k→∞ ln k
ln k
=
< 1 ⇒ original series converges
ln ( k + 1)
∑ ( −1)
k=2
ak+1
ak
28)
k
k2
( k + 1)
∑ ( −1) 10 k ⇒ ak+1 = 10 k+1 ≤ .01
k=4
4
k + 1 = 2 ⇒ 2 = .04 > .01
10
9
k + 1 = 3 ⇒ 3 = .009 < .01 ⇒ k = 2 ⇒ take 2 terms
10
i.e k = 4 and k = 5
∞
2
k
42
52
S2 = 4 − 5 = .00135 ≈ S
10 10
36)
∞
k!
; k = 11 ⇒ .00014 > .0001
k
k
k=1
k = 12 ⇒ .00005 < .0001
Hence, we need 11 terms with the 12th term
being the error bound.
∑ ( −1)
k+1
Page 663
∞
6)
∑ ( −1)
k=1
10)
k+1
k2 + 1
k2 + 1
; lim
= ∞ ⇒ original series diverges.
k→∞
k
k
∞
4
4
; lim
=0
2k + 1 k→∞ 2k + 1
4
2k + 1 2k + 1
=
⋅
=
<1
2 ( k + 1) + 1 4
2k + 3
∑ ( −1)
k=3
ak+1
ak
k+1
∴ original series converges by the Alternating
Series Test.
4
−4 ⋅ 2
−8
f ( x) =
⇒ f '( x) =
=
<0
2x + 1
( 2x + 1)2 ( 2x + 1)2
∴ f is decreasing. It is also continuous and positive.
∞
t
4
4
t
dx
=
lim
dx
=
lim
2
ln
2x
+
1
(
)
⎤
⎦3 =
∫3 2x + 1 t→∞ ∫3 2x + 1 t→∞
lim 2 ⎡⎣ ln ( 2t + 1) − ln 7 ⎤⎦ = ∞ ⇒
t→∞
∞
4
∑ 2k + 1 diverges ⇒
k=3
original series is conditionally convergent.
∞
16)
∞
4
1
=
4
∑ k ∑ k diverges ( harmonic series )
k=1
k=1
24)
∞
∞
cos k cos k
1
1
⇒ ∑ 3 ; 3 ≤ 3 ⇒ ∑ 3 is a convergent p − series.
k
k
k=1 k
k=1 k
Hence, the original series converges absolutely
by the Comparison Test.