HW SOLUTIONS WEEKS 56
1.
Disclaimer
Even though I’ve made every effort to make these solutions clear and readable, I
am only human and can make mistakes. If you find a mistake, please let me know
ASAP so I can correct it. Also, I have taken liberty of omitting some details that
aren’t necessary for me to repeat. However, if you don’t understand a step I have
omitted, ask for clarification.
2.
Week 5
1.
Suppose
f
n
→
f
uniformly on
S
. Let
>
0. Then
∃
N
∈
N
such that
n > N
=
⇒ 
f
n
(
x
)

f
(
x
)

<
/
2
∀
x
∈
S
Then for
m, n > N
, we get

f
n
(
x
)

f
m
(
x
)
 ≤ 
f
n
(
x
)

f
(
x
)

+

f
(
x
)

f
m
(
x
)

<
/
2 +
/
2 =
which implies
{
f
n
}
is uniformly Cauchy.
2.
Part (a). Let
x
∈
[0
,
1). Then
x
n
1+
x
n
≤
x
n
. The comparison test implies then
that
x
n
1 +
x
n
≤
x
n
=
1
1

x
Therefore the series converges.
Part (b). Let
a
∈
[0
,
1). For any
x
∈
[0
, a
], observe that
x
n
1 +
x
n
≤
a
n
1 +
x
n
≤
a
n
=
1
1

a
Therefore, by the Weierstrass Mtest (with
M
n
=
a
n
), the series converges uni
formly on [0
, a
].
Part (c).
The series does NOT converge uniformly on [0
,
1)!
Remember the
definition of uniform convergence of a series:
∑
∞
n
=1
g
n
→
g
uniformly if and only
if the sequence of partial sums
s
k
(=
∑
k
n
=1
g
n
)
→
g
uniformly. Then using Remark
24.4, we can say that
∑
g
n
→
g
uniformly if and only if
lim
n
→∞
sup
x
∈
S
{
g
(
x
)

s
n
(
x
)
}
= 0
In our case where
g
n
(
x
) =
x
n
1+
x
n
and
g
(
x
) =
∑
∞
n
=0
x
n
1+
x
n
, observe that
sup
x
∈
[0
,
1)
{
g
(
x
)

s
n
(
x
)
}
=
sup
x
∈
[0
,
1)
∞
k
=
n
+1
x
k
1 +
x
k
=
1
/
2
=
∞
Therefore the supremum is infinite for all
n
, so the limit is note 0, hence does not
converge uniformly.
3.
Part (a): Simply differentiate term by term.
Be careful to make sure you
index your summation correctly.
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 Fall '07
 Fukuda
 Calculus, Taylor Series, lim, Xn, xn xn lim

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