SOLUTION OF HW10
MINGFENG ZHAO
November 29, 2012
1. [10 Points] Suppose that ak and bk are nonnegative for all k ∈ N. Prove that if
converge, then
∞
X
∞
X
ak and
k=1
∞
X
bk
k=1
ak bk converges.
k=1
Proof. Since
∞
X
bk converges, then bk → 0 as k → ∞, which implies that there exists some M > 0
k=1
such that
|bk | ≤ M,
∀k ≥ 1
Since ak , bk ≥ 0 for all k ≥ 1, then
0 ≤ ak bk ≤ M bk ,
Since
∞
X
bk converges, by the comparision test, then
k=1
∀k ≥ 1
∞
X
ak bk converges.
k=1
2. [20 Points] For any sequence {xk }, define lim inf xk = lim
k→∞
a. Prove that if lim inf xk = lim
k→∞
inf xk . Then
n→∞ k>n
inf xk > x for some x ∈ R, then xk > x for large k.
n→∞ k>n
b. Prove that if xk → x for some x ∈ R, as k → ∞, then lim inf xk = x.
k→∞
c. Let ak > 0 for all k ∈ N, then
lim inf
k→∞
√
√
ak+1
ak+1
≤ lim inf k ak ≤ lim sup k ak ≤ lim sup
k→∞
ak
ak
n→∞
k→∞
d. Prove that if bn ∈ R\{0}, and
1
|bn+1 |
→ r for some r > 0 as n → ∞, then |bn | n → r as
|bn |
n → ∞.
1
2
MINGFENG ZHAO
Proof. For all n ∈ N, let
sn = inf xk ,
and tn = sup xk
k>n
k>n
Then we know that s1 ≤ s2 ≤ s3 ≤ · · · , t1 ≥ t2 ≥ t3 ≥ · · · , sn ≤ tn for all n ≥ 1,
lim sn = lim inf xk ,
n→∞
k→∞
a. Assume that lim inf xk = lim
k→∞
and
lim tn = lim sup xk
n→∞
k→∞
inf xk > x for some x ∈ R, that is,
n→∞ k>n
lim sn > x
n→∞
So there exists some N ∈ N such that for all n ≥ N , we have
sn > x
By the definition of sn , then
sk ≥ sn > x,
∀k ≥ N
b. Assume that xk → ∞x for some x ∈ R, as k → ∞, then for any > 0, there exists some K ∈ N
such that
x − < xk < x + ,
∀k ≥ K
Hence we know that for all n ≥ K + 1, we have
x − < xn = inf xk < x + k>n
By the definition of limit, we know that
lim xn = x
n→∞
Hence we get
lim inf xk = lim xn = x
k→∞
n→∞
SOLUTION OF HW10
3
c. By the definition of lim inf and lim sup, we know that
√
k
lim inf
k→∞
Claim I: lim inf
k→∞
If lim inf
k→∞
ak
√
ak+1
≤ lim inf k ak .
k→∞
ak
ak+1
= 0, it is easy to see that
ak
k→∞
k→∞
√
k
n→∞
0 = lim inf
If lim inf
ak ≤ lim sup
√
ak+1
≤ lim inf k ak
k→∞
ak
ak+1
ak+1
ak+1
> 0 or lim inf
= ∞, for any x > 0 such that x < lim inf
, by the result
k→∞
k→∞
ak
ak
ak
of part a, we know that there exists some K ∈ N such that
xk+1
> x,
xk
∀k ≥ K
On the other hand, we know that
xk
=
xk−1
xK+1
xk
·
···
· xK
xk−1 xk−2
xK
>
xk−K · xK ,
∀k ≥ K + 1
So we get
√
k
K
1
k
xk > x1− k · xK
,
∀k ≥ K + 1
Then we get
lim inf
k→∞
K
√
k
K
1
k
xk ≥ lim inf x1− k · xK
k→∞
1
k
Since lim x1− k · xK
= x, then
k→∞
K
1
k
lim inf x1− k · xK
=x
k→∞
Hence we get
lim inf
k→∞
√
k
xk ≥ x
4
MINGFENG ZHAO
Sinnce x > 0 is arbitrry smaller than lim inf
k→∞
√
ak+1
≤ lim inf k xk
k→∞
ak
lim inf
k→∞
√
k
ak+1
, then
ak
ak+1
.
ak
n→∞
k→∞
ak+1
For any x > 0 such that lim sup
< x, then there exists some K ∈ N such that
ak
k→∞
Claim II: lim sup
ak ≤ lim sup
ak+1
< x,
ak
∀k ≥ K
On the other hand, we know that
xk
=
xk
xk−1
xK+1
·
···
· xK
xk−1 xk−2
xK
<
xk−K · xK ,
∀k ≥ K + 1
So we get
√
k
1
K
k
xk < x1− k · xK
,
∀k ≥ K + 1
Then we get
lim sup
√
k
k→∞
K
K
1
k
xk ≤ lim sup x1− k · xK
k→∞
1
k
Since lim x1− k · xK = x, then
k→∞
K
1
k
lim sup x1− k · xK
=x
k→∞
Hence we get
lim sup
√
k
xk ≤ x
k→∞
Sinnce x > 0 is arbitrry greater than lim sup
k→∞
lim sup
k→∞
d. Assume that
√
k
ak+1
, then
ak
xk ≤ lim sup
k→∞
ak+1
ak
|bn+1 |
→ r for some r > 0, as n → ∞, then
|bn |
lim inf
n→∞
|bn+1 |
|bn+1 |
= lim sup
=r
|bn |
|bn |
n→∞
SOLUTION OF HW10
5
By the result of part c, then
lim inf
√
k
k→∞
√
k
ak = lim sup
ak = r
k→∞
For any > 0, then there exists some K ∈ N such that for all k ≥ N , we have
r−<
√
k
ak < r + By the definition of limit, then
lim
n→∞
√
k
ak = r
3.[10 Points] a. Prove that
b. Prove that
x
→ 0 uniformly as n → ∞, on any closed interval [a, b].
n
1
→ 0 pointwise but not uniformly on (0, 1) as n → ∞.
nx
Proof. a. For the intevral [a, b], then
|x| ≤ max{|a|, |b|} = M
For any > 0, let N =
M
+ 1, then
M
< N . So for all n ≥ N , we have
x |x|
M
≤
< ,
=
n
n
n
∀x ∈ [a, b]
x
→ 0 uniformly as n → ∞, on [a, b].
n
1
b. For any x ∈ (0, 1), since lim
= 0, then
n→∞ n
That is,
lim
n→∞
That is,
Claim I:
1
=0
nx
1
→ 0 pointwise, as n → ∞.
nx
1
does not uniformly converges to 0 on (0, 1) as n → ∞.
nx
Otherwise, then for any > 0, there exists some N ∈ N such that for all n ≥ N , we have
1 sup < nx
x∈(0,1)
6
MINGFENG ZHAO
Take = 1, for any n ∈ N, let xn =
Contradiction. Hence
1
, then
2n
1 nxn = 2 > 1
1
does not uniformly converges to 0 on (0, 1) as n → ∞.
nx
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu