# MATH 172 EXAM 3 Fall 1998 Section 502

```MATH 172
EXAM 3
Fall 1998
Section 502
Solutions
P. Yasskin
Multiple Choice: (5 points each)
1. Compute
a. 0
3n 2
lim
nv. 1 n 3
correctchoice
b. 1
c. 2
d. 3
3
3n 2 lim
n
0 0
lim
nv. 1 n 3
nv. 1
1
1
n3
e. Divergent
2. Find
5 5r 5r 2 5r 3 5r 4 such that
r
a. 2
C 3.
5
b. &quot; 2
c. 3
5
5
d. 5
3
e. &quot; 2
5 3
1&quot;r
correctchoice
3
!
.
3. The series
1
n2 n
n1
5 1&quot;r
3
r 1&quot; 5 &quot;2
3
3
is
!
!
.
a. divergent by comparison to
n1
b. convergent by comparison to
1 .
n
.
n1
1 .
n2
correctchoice
c. divergent by the ratio test.
d. convergent by the ratio test.
e. divergent by the n th -term test.
!
.
In the series
n1
1
n2 n
the largest term in the denominator is
p-series since
!
.
So
n1
p 2 1.
1
n2 n
!
.
apply the Comparison Test by comparing with
n1
Since
n2 n n2
1
n2
n2.
So, we
which is a convergent
we have
1
12 .
n
n2 n
is also convergent.
1
!
99
4. Compute
1
k
correctchoice
k1
a. .9
b.
c.
d.
e.
1
&quot;
k1
.99
1
1.1
Divergent
!
99
k1
1
1
&quot;
k1
1 &quot; 1 .9
10
k
!
.
5. Compute
n1
1
&quot;
1
1
2
1
&quot;
2
1
2
C
1
&quot;
99
1
100
3n 2
1 n3
a. ln 2
b. 3
2
c. 27
82
d. Convergent but none of the above
correctchoice
e. Divergent
;

.
3n 2 dn ln 1 n 3
3
1 1n
the Integral Test.
!
.
6. The series
n1
&quot;1
.
1
. &quot; ln 2 .
!
.
So
3n 2
1 n3
n1
is divergent by
n
is
3 n
a. absolutely convergent.
b. conditionally convergent.
correctchoice
c. absolutely divergent.
d. conditionally divergent.
e. oscillatory divergent.
!
.
The series
n1
series and
lim
nv.
&quot;1
n
is convergent because it is an alternating, decreasing
3 n
1
3 n
0.
!
.
The related absolute series is
which is a divergent p-series since
p 1 1.
2
!
.
So
n1
n1
&quot;1
3 n
1
3 n
1
3
!
.
n1
1
n 1/2
n
is
conditionally convergent.
2
7. Compute
lim
xv0
 cos 2x &quot; 1 2x 2
x4
a. 0
1
24
c. 1
12
d. 2
3
b.
correctchoice
 cos 2x &quot; 1 2x 2
lim
lim
xv0
xv0
x4
e. .
2x &quot; C
4!
2x 1&quot;
2
2!
2x &quot; C
4!
4
&quot; 1 2x 2
x4
4
lim
xv0
!
.
8. Given that
1&quot;x
1
1&quot;x
x
c.
1&quot;x
x
d.
1&quot;x
n
e.
1&quot;x
b.


(for |x| 1),
1&quot;x
.
n xn n0
!
xn 1 .
1&quot;x
to get
x
!
.
nx n !
.
n0
1
n!
1
2
n
1 &quot; x !
.
d
dx
Apply
n0
9. The series
!
then (for |x| 1) we have
correctchoice
2
n0
Multiply by
3
2
.
16 2
24
1
xn n0
1
a.
x4
to get
n0
x
2
nx n&quot;1 1 &quot; x 1
2
.
.
converges to
a. ln 2
b.
correctchoice
e
c. sin 1
2
d. sin 2
e. e 2
 !
.
n0
 1 x
n!
n
e
!
.
x
So:
n0
1
n!
1
2
n
e 1/2 e
3
10. Find the
3 rd degree
x 2.
a. 3 x &quot; 2 3
8
b. &quot;3 x &quot; 2 3
8
c. &quot;6 x &quot; 2 3
d. &quot;1 x &quot; 2 3
16
e. 1 x &quot; 2 3
16


 
term in the Taylor series for
 f x 1x
centered at
  correctchoice
 f U x &quot;12
f UU x 23
x
x
rd
So the 3 degree term is
f
f
UUU
x &quot;x6
2 x &quot; 2 UUU
3!
4
3
f
UUU
2 1 &quot;3
6
8
&quot;6 &quot; 3
8
16
x &quot; 2 3

&quot; 1 x &quot; 2 3.
16
4
!
.
11. (15 points) Find the interval of convergence for the series
n1
x&quot;5 n
.
3nn3
Be sure to identify each of the following and give reasons:
a
(1 pt) Center of Convergence:
an x &quot; 5 n
a n1 3nn3
a n1
&gt; lim
an
nv.
5
x &quot; 5 3 n 1 n1
n1
3
x &quot; 5 3 n
3 n 1 x &quot; 5 n1
nlim
v.
n1
n 3
3
lim
nv.
n
|x &quot; 5|
1
3
The series converges if
or
x &quot; 5 3
n
n1
x
! 8 &quot; 5 .
The series
n1
p 3 1.
n
3nn3
&quot; 5|
|x
3
|x &quot; 5| 3
(1 pt) Right Endpoint:
3
R
3
(5 pt)
53 8
!
.
n1
1
n3
converges because it is s p-series with
At the Right Endpoint the Series
Converges
(circle one)
(3 pt)
Diverges
! 2 &quot; 5 .
The series
!
.
series
n1
and
n1
1
n3
5&quot;3 2
x
(1 pt) Left Endpoint:
n 3
3 n
n
!
.
n1
converges
&quot;1
n
n
3
OR
converges because its related absolute
because it is an alternating decreasing series
lim 1 0.
nv. n 3
At the Left Endpoint the Series
Converges
(circle one)
(3 pt)
Diverges
(1 pt) Interval of Convergence:
2txt8
or
&iexcl;2, 8 &cent;
5
 12. (15 points) Let
f x x 2 cos x.
a. (10 pt) Find the Maclaurin series for
 fx .
and also write out the first 4 terms.
cos x !  !  .
&quot;1
x 2n 1 &quot; x 2 x 4 &quot; x 6 2n !
2
4!
6!
n
n0
.
x cos x 2
&quot;1
n
n0
Write the series in summation form
C
x 2n2 x 2 &quot; x 4 x 6 &quot; x 8 2n !
2
4!
6!
 C
f 6 0 .
b. (5 pt) Find
x6
The term with an
 6
f 6 0 6
x x .
4!
6!
is
C

  f 6 0 6! 30.
4!
So:
find the 6 th degree Taylor
ln 1 t t &quot; 1 t 2 1 t 3 &quot; ,
3
2
x0
for
ln 1 x 2 .
13. (10 points) Given that

ln 1 x 2 x 2 &quot; 1 x 4 1 x 6
3
2
t x2:
Just substitute
14. (15 points) You are given:
e
;
&quot;x 2
!
.
&quot;1
n0
n
&quot;C
x 2n 1 &quot; x 2 x 4 &quot; x 6 n!
2
6
x0
&quot;x 2
e
to estimate
0.1
e
&quot;x 2
dx.
0.1
e
&quot;x 2
0
dx ;
0.1
0
(Keep 8 digits.)
3
1 &quot; x 2 dx x &quot; x
3
0.1
0
Since
e
&quot;x 2
and
most the next term:
;
0.1
0
x 4 dx 2
x5
10
0.1
0
;
0.1
e
&quot;x 2
dx
e
&quot;x 2
1 &quot; x2.
So
.1 &quot; .001 .09966667
b. (5 pt Extra Credit) Your result in (a) is equal to
much? Why?
for
0
The quadratic Taylor polynomial approximation is
integrate:
;
C.
;
3
0.1
0
e
&quot;x 2
dx
to within o how
are alternating decreasing series, the error is at
0
.1 10
5
10 &quot;6 .000001
6
```