Math 521  Advanced Calculus I
Instructor: J. Metcalfe
Due: February 12, 2010
Assignment 10
1.
Let
{
x
n
}
be a bounded sequence, and for each
n
∈
N
let
s
n
:= sup
{
x
k
:
k
≥
n
}
and
t
n
:=
inf
{
x
k
:
k
≥
n
}
. Prove that
{
s
n
}
and
{
t
n
}
are monotone and convergent. Also prove that if
lim
s
n
= lim
t
n
, then
{
x
n
}
is convergent. [One calls lim
s
n
the
limit superior
of
{
x
n
}
, lim sup
x
n
,
and lim
t
n
the
limit inferior
of
{
x
n
}
, lim inf
x
n
.]
Since
{
x
n
}
is bounded, we know that
s
n
and
t
n
exist for each
n
. Moreover, we have
that
X
n
+1
:=
{
x
k
:
k
≥
n
+ 1
} ⊂ {
x
k
:
k
≥
n
}
=:
X
n
.
Thus any upper bound for
X
n
is automatically an upper bound for
X
n
+1
. In particular
s
n
is an upper bound for the set
X
n
+1
. And since
s
n
+1
is its least upper bound, it
follows that
s
n
+1
≤
s
n
.
Similarly, a lower bound for
X
n
, namely
t
n
, is automatically a lower bound for
X
n
+1
.
Since
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 Spring '10
 MetCalfe
 Calculus, Tn, Supremum, Limit of a sequence, Limit superior and limit inferior, Xn, subsequence

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