Math 5C: Exam #2 Solutions
Date:
July 16
th
, 2010
Score: out of 60
1.
(10)
Match each Maclaurin series to the function from the following list it represents
by ﬁlling in the blank space below the series. (Note: All listed function are
C
∞
at
x
= 0
under the convention
f
(0) := lim
x
→
0
f
(
x
)
.)
(i)
∞
∑
k
=0
(

1)
k
x
2
k
=
∞
∑
k
=0
(

x
2
)
k
=
1
1

(

x
2
)
=
1
1+
x
2
(ii)
∞
∑
k
=0
(

1)
k
k
!
x
2
k
=
∞
∑
k
=0
(

x
2
)
k
k
!
=
e

x
2
(iii)
∞
∑
k
=0
(

1)
k
(2
k
)!
x
2
k
= cos(
x
)
(iv)
∞
∑
k
=0
(

1)
k
(2
k
+1)!
x
2
k
=
1
x
∞
∑
k
=0
(

1)
k
(2
k
+1)!
x
2
k
+1
=
1
x
sin(
x
)
(v)
∞
∑
k
=0
(

1)
k
2
k
+1
x
2
k
=
1
x
∞
∑
k
=0
(

1)
k
2
k
+1
x
2
k
+1
=
1
x
arctan(
x
)
2.
(5+5)
Use appropriate convergence tests to discern the convergence (absolute,
conditional, etc.) or divergence of the following series. Identify
which tests you are using.
(2a)
Let
∞
∑
k
=1
u
k
=
∞
∑
k
=1
(

1)
k
k
+
√
k
. As
{
u
k
}
∞
n
=1
is a monotone decreasing sequence that
converges to zero, by the Alternating Series Test, the series converges. For
∞
∑
k
=1

u
k

, consider
the Limit Comparison Test to the harmonic series
∞
∑
k
=1
1
k
. As
lim
k
→∞
1
k

u
k

= lim
k
→∞
k
+
√
k
k
= 1
>
0
,
the original series converges absolutely if and only if the harmonic series converges. As the
harmonic series diverges,
∞
∑
k
=1
(

1)
k
k
+
√
k
converges conditionally
.
Note:
Using the direct Comparison Test (
∞
∑
k
=1

u
k
 ≥
∞
∑
k
=1
1
2
k
) also shows absolute divergence.
(2b)
Let
∞
∑
k
=1
u
k
=
∞
∑
k
=1
sin
k
(
1
k
)
. As
sin
k
(
1
k
)
>
0
for
k
∈
N
, by the Root Test, as
lim
k
→∞
(
sin
k
(
1
k
)
)
1
k
= lim
k
→∞
sin(
1
k
) = sin(0) = 0 =
ρ <
1
, the series
converges
(absolutely).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document3.
(5+5)
Let
f
(
x
) =
sin
(2
x
)
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '06
 Roche
 Maclaurin Series, Vector Calculus, Mathematical Series, UK

Click to edit the document details