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MATH 2300 - Fall 2008
Exam 3 Solutions
1. (a)
1X
k=2
34
k+1
=
34
3 1X
k=2
34
k 2
= 2764 11 + 3
4
= 2764 47 = 27112
(b)
nX
k=0
tan 1(k+1) tan 1(k) =
n+1X
k=1
tan 1(k)
nX
k=0
tan 1(k) = tan 1(n+1) tan 1(0) = tan 1(n+1):
1X
k=0
tan 1(k+ 1) tan 1(k) = lim
n!1
nX
k=0
tan 1(k+ 1) tan 1(k) = lim
n!1tan
1(n+ 1) =
2:
2. (a) lim
k!1
1
1=k = 16= 0: So,
1X
k=1
1
1=k diverges by the Divergence Test.
(b) Since fkg and flnkg are monotone increasing to in nity, then f 1klnkg is monotone de-
creasing to zero. So, by the Alternating Series Test,
1X
k=2
( 1)k
klnk converges.
Also, x and lnx are continuous functions and nonzero for x> 0: So, 1xlnx is also contin-
uous. Furthermore, it is decreasing since x and lnx are increasing. Note that
Z 1
2
1
xlnxdx =
Z 1
ln2
1
udu = lima!1[lnu]
a
ln2 = lima!1lna ln(ln 2) =1:
So, by the Integral Test,
1X
k=2
1
klnk diverges. Thus,
1X
k=2
( 1)k
klnk converges conditionally.
3. (a)
1X
k=1
4 + ( 1)k
...

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