MATH 172
EXAM 3
Section 504
Spring 1999
Solutions
P. Yasskin
Multiple Choice: (6 points each)
lnn
lim
nv. n
1. Compute
a. ".
b. "1
c. 0
correctchoice
d. e
e. .
By l’Hopital’s Rule:
!
.
2. The series
1
lnn lim n
lim
nv. n
nv. 1
lnn
n
n1
0
is
a. Divergent by the n th Term Divergence Test
b. Divergent by the Integral Test
correctchoice
c. Convergent by the Integral Test
d. Divergent by the Ratio Test
e. Convergent by the Ratio Test
1
lnn lim n 0
lim
nv. n
nv. 1
ln n is decreasing and . lnn dn n
n
1
n th Term Divergence Test:
Integral Test:
!
.
So
n1
;
lim aan1
n
nv.
nlim
v.
lnŸn 1 n
1 lnŸn Ÿn
1
3. A convergent sequence is recursively defined by
a. 19
a .
lim
nv. n
24
21 " 3
b.
2
c.
0
u du u2
2
.
0
.
lnn diverges.
n
Ratio Test:
Find
;
Test Fails.
.
Test Fails.
a1 1
and
a n1 3 " a n .
2 an
correctchoice
=
4
d. 3
2
e. 2
3
1
a . Then the limit of the recurrsion relation says L 3 " L . Thus
Let L nlim
v. n
2L
"
3
o
9
"
4Ÿ"3 or
L 2 3L " 3 0
or
L
"3 o 21
2L L 2 3 " L
2
2
!
.
4. The sum of the series
Ÿ" 1 n1 n1
3
4
n1
n
is:
a. nonexistant since its partial sums oscillate
b. " 3
4
9
c.
correctchoice
7
d. nonexistant by the n th Term Divergence Test
e. 3
16
This is a geometric series with ratio r " 3 and first term a 4
9
.
n1 n1
3
Ÿ" 1 4
So
9 9
n
7
43
4
3
1""
n1
4
!
!
.
5. Compute
k2
a. 9
3
Ÿn
" 1 Ÿn 1 Ÿ" 1 4
2 2
3
1
9.
4
. (HINTS: partial fractions,telescoping sum)
correctchoice
4
b. 3
2
c. 1
d. 1
2
e. .
3
Partial Fractions:
n1:
3 AŸ2 3
Ÿn
Ÿn
" 1 Ÿn 1 3
2
" 1 Ÿn 1 A 3
2
1 " 1
n"1
n1
A B
3 AŸn 1 BŸn " 1 n"1
n1
So
3 BŸ"2 n "1 :
B "3
2
.
.
3
1 " 1
and
3
2
"
1
n
n1
Ÿn " 1 Ÿn 1 k2
k2
!
!
The partial sum is
!
k
Sk 3
2
Thus
1 " 1
"
1
n
n1
k2
1 " 1 1 " 1 1 " 1 C
1 " 1
3
5
2
1
2
3
k"2
3
4
k
1
1
1
3
1
" "
k
k1
2 1
2
.
3
nlim
S 3 1 1 9
v. k
2 1
2
4
Ÿn " 1 Ÿn 1 k2
1 " 1
k"1
k1
!
2
!
.
6. The series
Ÿ" 1 n1
n
3nn2
n!
is
a. Absolutely convergent
correctchoice
b. Conditionally convergent
c. Divergent
n 2
3 n1 Ÿn 1 2
a n Ÿ" 1 n 3 n
a n1 Ÿ"1 n1
n!
Ÿn 1 !
2
n1
1
a n1 lim 3 Ÿn 3
n!
n1 2 0
nlim
L nlim
n
2
n
v. a n
nv.
v.
n1
Ÿn 1 !
3 n
So the related absolute series converges and hence the original series is absolutely
convergent.
Ratio Test:
!
.
7. The series
n1
Ÿ" 1 n Ÿn
3 2 2n
is
3 n100
a. Absolutely convergent
b. Conditionally convergent
c. Divergent
correctchoice
n th Term Divergence Test:
lim
nv.
Ÿn
3 2 2n
3 n100
a n1 lim Ÿn 4 2 2Ÿn1 Ratio Test:
lim
an
nv.
nv.
3 n101
In either case the series is divergent.
2
3
4
lnŸ1 x x " x x " x C ,
2
3
4
lnŸ1 2x " 2x 2x 2
3
Ÿ2x 0
8
3
4
3
1
correctchoice
3
8. Given that
lim
xv0
a.
b.
c.
d.
1 limŸn 3 4 n .
OR
3
3 100 nv.
3 n100
4 Ÿn 4 4 1
nlim
2n
v.
3
3 Ÿn 3 3
2
n
Ÿ
compute
e. .
2
3
4
lnŸ1 2x 2x " 4x 8x " 16x C
2
3
4
2
2x " 4x lnŸ1 2x " 2x 2x 2
2
lim
lim
3
xv0
xv0
Ÿ2x 3
8x " 16x 4 C
3
4
lim
lim
xv0
xv0
8x 3
8x 3 " 16x 4 C " 2x 2x 2
3
4
3
Ÿ2x 8 " 16x C
3
4
1
3
8
3
!
.
9. For what values of x does the series
x
x
0
x
e. x
a.
b.
c.
d.
4 only
3 only
x 3 only
1 or x 3 only
2 or x 4 only
n0
1
Ÿx
" 3 n
converge?
correctchoice
This is a geometric series with ratio
r 1 .
It converges when |r| 1, or
x"3
1
1
or
1 |x " 3|.
This means x " 3 "1
or
x " 3 1.
x"3
Equivalently, x 2 or x 4.
10. In the Maclaurin series for
sinŸx 2 x
the coefficient of
x9
is
a. 1
1
3!
c. 1
5!
d. 1
7!
e. 1
9!
b.
correctchoice
3
5
7
we have
sin x x " x x " x C
3!
5!
7!
6
10
14
5
9
13
sinŸx 2 sinŸx 2 x 2 " x x " x C
and
x" x x " x C.
x
3!
5!
7!
3!
5!
7!
1 .
So the coefficient of
x9
is
5!
Since
4
11. (25 points) You are given:
cosŸx 2 fŸx cosŸx 2 ,
a. (5 pt) If
x6
The term with an
Ÿ" 1 n0
Ÿ6 find
f
n
x 4n 1 " x 4 x 8 " x 12 C.
2!
4!
6!
Ÿ2n !
Ÿ0 .
f Ÿ6 Ÿ0 6
x 0.
6!
is
fŸx cosŸx 2 ,
b. (5 pt) If
!
.
f Ÿ6 Ÿ0 0.
So:
f Ÿ16 Ÿ0 .
find
16
f Ÿ16 Ÿ0 16
x x .
8!
16!
16!
Ÿ16 16 15 14 13 12 11 10 9 518 918 400.
f Ÿ0 8!
x 16
The term with an
So:
is
;
c. (10 pt) Use the quartic (degree 4) Taylor polynomial approximation about
for
cosŸx
2
to estimate
0.1
0
cosŸx
2
dx.
The quartic Taylor polynomial approximation is
;
0.1
0
cosŸx 2 dx X
;
0.1
0
4
1 " x dx 2
d. (5 pt) Your result in (c) is equal to
Why?
Since
;
cosŸx 2 and
;
0.1
0
0.1
5
x" x
10
;
0.1
0
0
4
cosŸx 2 X 1 " x .
2!
So:
5
.1 " .1 .099999
10
cosŸx 2 dx
cosŸx 2 dx
x0
to within o how much?
are alternating decreasing series, the
error is at most the next term:
0.1
0.1 8
.1 9
x dx x9
Ÿ 4.6 • 10 "12
4!
9 4! 0
9 24
0
5
!
.
12. (15 points) Find the interval of convergence for the series
n0
Ÿx
2
" 6 n .
n
n
Be sure to identify each of the following and give reasons:
a
(1 pt) Center of Convergence:
an Ÿx
" 6 n
a n1 2n n
a n1
> lim
an
nv.
nlim
v.
Ÿx "
2 n1
The series converges if
Ÿx "
2 n1
6 n1
n1
6
6 n1
n1
2n n
n
Ÿx " 6 |x " 6|
1
2
lim
nv.
or
Ÿx
" 6 2
n
n1
!
!
.
n1
2
|x " 6| 2
Radius of Convergence:
(1 pt) Right Endpoint:
x
.
n
Ÿ8 " 6 The series
2n n
n1
p 1 1.
2
" 6|
|x
R
2
(5 pt)
62 8
1
n
diverges because it is s p-series with
Converges
At the Right Endpoint the Series
(circle one)
(3 pt)
Diverges
6"2 4
n
Ÿ" 1 converges because it is an alternating
n
n1
1 0.
lim
nv.
n
(1 pt) Left Endpoint:
x
.
n
Ÿ4 " 6 The series
2n n
n1
!
decreasing series and
!
.
At the Left Endpoint the Series
Converges
(circle one)
(3 pt)
Diverges
(1 pt) Interval of Convergence:
4tx8
or
¡4, 8 6