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55.
Given an
e
.
0, by definition of convergence there
corresponds an
N
such that for all
n
,
N
,
)
L
1
2
a
n
)
,
e
and
)
L
2
2
a
n
)
,
e
. (There is one such number for each series,
and we may let
N
be the larger of the two numbers.) Now
)
L
2
2
L
1
)
5
)
L
2
2
a
n
1
a
n
2
L
1
)
#
)
L
2
2
a
n
)
1
)
a
n
2
L
1
)
,
e
1
e
5
2
e
.
)
L
2
2
L
1
)
,
2
e
says that the difference between two fixed
values is smaller than any positive number 2
e
. The only
nonnegative number smaller than every positive number
is 0, so
)
L
2
2
L
1
)
5
0 or
L
1
5
L
2
.
56.
Consider the two subsequences
a
k
(
n
)
and
a
i
(
n
)
, where
lim
n
→
‘
a
k
(
n
)
5
L
1
,lim
n
→
‘
a
i
(
n
)
5
L
2
, and
L
1
±
L
2
. Given an
e
.
0
there corresponds an
N
1
such that for
k
(
n
)
.
N
1
,
)
a
k
(
n
)
2
L
1
)
,
e
, and an
N
2
such that for
i
(
n
)
.
N
2
,
)
a
i
(
n
)
2
L
2
)
,
e
. Assume
a
n
converges.
Let
N
5
max {
N
1
,
N
2
}. Then for
n
.
N
, we have that
)
a
n
2
L
1
)
,
e
and
)
a
n
2
L
2
)
,
e
for infinitely many
n
. This
implies that lim
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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