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1
Solutions to Exam #1  Version
α
1)
a)
n=1
1
n(n + 5)
∞
∑
converges
by the comparison test
since
2
11
n(n
5)
n
≤
+
and
∑
∞
=
1
n
2
n
1
converges by the integral test.
b)
2
n
2
n=1
n+
1
(1)
2n +1
∞
∑
diverges
since the terms don’t approach 0 as
n goes to infinity.
The absolute value of a term is
2
2
1
21
n
+
+
which approaches 1/2
by l’Hôpital’s rule and thus the actual
term has no limit.
Then even terms go to 1/2 and the odd ones
to 1/2.
c)
1
(2 )!
∞
=
∑
converges by
comparison
with
e
=
∑
∞
=
1
n
n!
1
,
since
!
1
!
1
)
1
(
)
1
2
)(
2
(
n
n
n
n
n
n
n
n
≤
+
−
⋅
=
"
"
(2n)!
n
n
.
e
=
∑
∞
=
1
n
n!
1
converges by the ratio test as
1
!0
(1
)
!
1
a
an
+
==
→
++
i
,
as n goes to infinity.
You could also use the ratio test on the original series,
but that is a little harder.
2)
f(x)=
n
n
n
n=1
x
(1)
5
∞
∑
is a
geometric series.
If we substitute
u=x/5,
then
f(x)=
n
n=1
1
u=u
1u
∞
∑
,
converging for
u<1.
a)
The radius of convergence
of the series for f(x) is
R=5,
since we need
u=x/5<1
or
x<5.
Or we can use the
ratio test to find the radius of convergence.
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 Spring '06
 Mohanty

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