Spring 2006 Math 152
Exam 3A: Solutions
c
Mon, 01/May
!2006,
Art Belmonte
Thus A + B = 0 and A = 1, whence B = −A = −1.
%
∞
∞ $
!
!
1
1
1
Therefore,
=
−
.
n (n + 1)
n
n+1
n=1
• Now look at the sequence of partial sums to determine
its limit s, the sum of the series.
Notes
s1
• The notation bn ↓ 0 means 0 < bn+1 ≤ bn and lim bn = 0;
i.e., the sequence {bn } decreases to zero in the limit.
• GFF [“Generalized Fun Fact”]: Let p (n) =
m
!
s2
ck n k be a
k=0
"
n
p (n) = 1. This
n→∞
boosts the power of the Root Test, which was optionally
covered in some classes. (In most classes, only the Ratio Test
was covered.)
polynomial in n with cm > 0. Then lim
sn
s = lim sn
n→∞
) (−1)n
√
4n
) (−1)n
• I. The series
converges absolutely since
8
)
) 1 n
|an | =
,
a
convergent
p-series ( p = 8 > 1).
n8
'
(
)
5
• II. The series (−1)n+1 n n+2
diverges by the Test
3
for Divergence since lim an (= 0. (Indeed, the limit does
not exist since lim sup an = ∞ and lim inf an = −∞.)
) (−1)n
√
• III. The alternating series
converges by the
4
∞ n
∞ $ %n
∞ $ % $ %n−1
!
!
!
2
2
a
2/3
2/3
2
2
=
=
=
=
=
=2
2
3n
3
3
3
1−r
1/3
1
−
3
n=1
n=1
n=1
7. (d) The series
1
√
4n
∞
!
n=3
-
∞ '
∞
(n !
!
1
4
*
+=
=
−x
(−1)n x 4n ,
1 − −x 4
n=0
n=0
,
,
provided ,−x 4 , < 1 or |x| < 1.
∞
3
1
dx
x ln x
3. (b) We have
4. (d) The series
n=1
1
n+1
6. (d) Compute the sum of this geometric series via the
Geometric Series Theorem.
Alternating Series Test since bn = |an | =
↓ 0.
)
) 1
Note, however, that |an | =
is a divergent
n 1/4
) (−1)n
1
√
p-series ( p = 4 ≤ 1). Accordingly,
4n
converges, but is not absolutely convergent. (We say it
is conditionally convergent.)
∞
!
..
.
= 1−
= 1
1
3
)
• I. If lim bn = 0, then bn converges. This
false.
) is
1
A counterexample is the harmonic series
,
n
a divergent p-series ( p = 1 ≤ 1).
)
)
• II. If 0 ≤ an ≤ bn and bn diverges, then) an
diverges. This is false. A countexample is an = n12
)
)1
)1
and bn =
n . Note that
n , the harmonic series,
1
diverges. Yet while 0 ≤ an = n 2 ≤ n1 = bn , we see that
) 1
converges ( p-series with p = 2 > 1).
n2
)
• III. If an converges, then lim an = 0. This is true.
As stated in Section 10.2, this condition is necessary for
a series to converge.
converges, but not absolutely.
n
= 1 − 21
( '
(
'
=
1 − 12 + 21 − 31 = 1 −
5. (c) Only statement III is true.
1. (b) Algebraic manipulation gives
#
$
%&
πn
n2 + 1
lim
sin
n→∞
2n + 1
n2
#
#
&&
1 + n12
π
= lim
sin
n→∞
1
2 + n1
'π (
= sin
= 1.
2
2. (d) Only the series
n=1
1
diverges by the Integral Test.
n ln n
t
1 1
dx
ln x x
- ln t
1
= lim
du (sub u = ln x)
t→∞ ln 3 u
,ln t
,
= lim ln u ,
=
lim
t→∞ 3
t→∞
=
1
is a telescoping sum.
n (n + 1)
-
ln 3
lim (ln (ln t) − ln (ln 3)) = ∞
t→∞
$
%
1
sin n
1
1
8. (d) Note that − ≤
≤ . Since lim −
= 0 and
n→∞
n
n
n
n
1
sin n
lim
= 0, we conclude that lim
= 0 by the
n→∞ n
n→∞ n
Squeeze Theorem.
• First do a partial fraction decomposition.
1
A
B
=
+
n (n + 1)
n
n+1
1 = A (n + 1) + Bn
0n + 1 = (A + B) n + A
9. (c) The series
∞ 2n
!
x
√ converges for −1 < x < 1 as follows.
n
n=1
1
• The Ratio Test gives
,
,
, an+1 ,
,
lim ,,
=
n→∞ an ,
∞
!
(x − 3)n
has radius of convergence
n 4n
n=1
R = 4 and interval of convergence I = [−1, 7).
11. The power series
√ &
|x|2n+2
n
lim √
2n
n→∞
n + 1 |x|
.
1
|x|2
lim
n→∞ 1 + 1
n
#
=
• The Ratio Test gives
,
,
, an+1 ,
,
,
lim
n→∞ , an ,
need
|x|2 < 1
|x| < 1.
=
=⇒
• [Alternatively, use the Root Test along with the GFF.
As n → ∞, we have
* 2n +1/n
"
|x|
|x|2
need
n
|an | = "
= " √ → |x|2 < 1,
n √
n
n
n
whence |x| < 1 and R = 1.]
• At the endpoints x = ±1, the series is
1
2
)
1
,
n 1/2
=
=⇒
n=1
∞
!
n=1
• [Alternatively, use the Root Test along with the GFF.
As n → ∞, we have
a
"
|x − 3|
|x − 3| need
n
|an | = √
< 1,
=
n
4
4 n
whence |x − 3| < 4 and R = 4.]
• At the left endpoint x = −1, we have the alternating
) (−1)n
series
n , which converges by the Alternating
Series Test since bn = |an | = n1 ↓ 0.
∞
!
1
1
=
p
p
(1 + n) n
n + n p+1
n=1
• At the )
right endpoint x = 7, we have the harmonic
1
series
n , a divergent p-series ( p = 1 ≤ 1).
converges for p > 0 by the Limit Comparison Test.
• Let
!
bn =
!
• So the interval of convergence is I = [−1, 7).
1
. Then
p+1
n
an
1
n p+1
lim
= lim
= 1 > 0.
= lim
p
n→∞ bn
n→∞ 1 + 1
n→∞ (1 + n) n
n
! 1
Since
converges only if q = p + 1 > 1 or
n p+1
p > 0, we conclude by the Limit Comparison Test that
!
1
converges only if p > 0. [Brian Winn]
(1 + n) n p
• Here is an alternative proof that employs repeated use
of the Comparison Test instead.
12.
∞
∞
!
!
1
1
1
– First,
≤
=
,
p
p+1
p+1
nq
n +n
n
n=1
n=1
n=1
which converges provided p + 1 = q > 1 or
∞
!
1
converges for
p > 0. Therefore,
(1 + n) n p
n=1
p > 0 by the Comparison Test.
– Next,
n=1
1
n p + n p+1
≥
∞
!
n=1
1
n p+1 + n p+1
=
∞ n
!
e
converges absolutely (and hence
(a) The series
n!
n=1
converges) by the Ratio Test.
#
&
,
,
n+1
, an+1 ,
e
n!
, = lim
lim ,
·
n→∞ (n + 1)! en
n→∞ , an ,
e
= lim
=0<1
n→∞ n + 1
∞
!
(−1)n
converges by the
(b) The alternating series
1 + ln n
n=1
, ,
, ,
1
Alternating Series Test since bn = , a1n , = 1+ln
n ↓ 0.
∞
!
∞
!
|x − 3| need
< 1
4
|x − 3| < 4.
Hence the radius of convergence is R = 4.
10. (a) The series
an =
&
n
=
divergent p-series ( p = ≤ 1). So the interval of
convergence is I = (−1, 1).
∞
!
|x − 3|n+1
n4n
lim
n→∞ (n + 1) 4n+1 |x − 3|n
&
#
|x − 3|
1
lim
n→∞ 1 + 1
4
=
Hence the radius of convergence is R = 1.
#
(c) The series
n=1
n
converges by the Comparison
n 2.001 + 4
0≤
n 2.001
Theorem since
∞
∞
1 ! 1
1 ! 1
,
=
p+1
2
2
nq
n
n=1
∞
!
n=1
and
which diverges for p + 1 = q ≤ 1 or p ≤ 0.
∞
!
1
Therefore,
diverges for p ≤ 0 by
(1 + n) n p
n=1
the Comparison Test.
∞
!
1
converges only when
– Accordingly,
(1 + n) n p
n=1
p > 0.
)
1
n 1.001
n
+4
≤
n
n 2.001
=
1
n 1.001
is a convergent p-series ( p = 1.001 > 1).
13. Let f (x) = e3x and a = 2. Then f (n) (x) = 3n e3x and
f (n) (2) = 3n e6 . So the Taylor series for f (x) = e3x at
a = 2 is
∞
∞ n 6
!
!
f (n) (2)
3 e
(x − 2)n =
(x − 2)n .
n!
n!
n=0
2
n=0
14.
(a) Recall that sin z =
- 1
0
'
(
sin x 2 d x
=
=
=
=
=
∞
!
(−1)n z 2n+1
for z ∈ R. Thus
(2n + 1)!
n=0
' (2n+1
- 1!
∞ (−1)n x 2
0 n=0
- 1!
∞
0 n=0
(2n + 1)!
dx
(−1)n x 4n+2
dx
(2n + 1)!
,x=1
,
n x 4n+3
,
(−1)

,
(2n + 1)! (4n + 3) ,,
n=0
x=0

 

∞
∞
n 14n+3
!
!
(−1)
−

0
(2n + 1)! (4n + 3)

∞
!
n=0
n=0
∞
!
n=0
(−1)n
(2n + 1)! (4n + 3)
(b) The Alternating Series Estimation Theorem gives
|R9 | ≤ |a10 | =
15.
1
1
=
.
2
100
10
(a) One unit vector that is parallel to v = [1, 2, −2] is
3
4
v
1 2
2
[1, 2, −2]
v̂ =
=√
=
, ,− .
,v,
3 3
3
1+4+4
4
3
2 2
1
Another is −v̂ = − , − , .
3
3 3
(b) The vector (parallel) projection of b = [4, 2, 0] onto
a = [1, −1, 1] is
$
%
a·b
a
proja b =
,a, ,a,
$
%
4−2+0
[1, −1, 1]
=
√
√
1+1+1
3
3
4
2
2
2 2
=
,− ,
.
[1, −1, 1] =
3
3
3 3
3