Homework 1 Solutions
Chapter 2, Exercise 12
We prove that if sn converges to s, then
1
n
∑n
1 sk also converges to s. So suppose
sn → s. Then for any � > 0, we can choose N = N(�) so that |sn − s| < � if n ≥ N . Then we define C := ∑N 1 |sk − s|. Then for n > N we may write∣∣∣∣ 1n
n∑
k=1
sk − s
∣∣∣∣ =
∣∣∣∣ 1n
n∑
k=1
(sn − s)
∣∣∣∣
≤
1
n
n∑
k=1
|sn − s|
=
1
n
N∑
k=1
|sn − s|+
1
n
n∑
k=N+1
|sn − s|
≤
C
n
+
1
n
n∑
k=N+1
�
<
C
n
+ �.
Letting N →∞ on both sides, we have
lim sup
N→∞
∣∣∣∣ 1n
n∑
k=1
sk − s
∣∣∣∣ ≤ �,
which proves the claim since � is arbitrary.
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Chapter 2, Exercise 13
Part a: It suffices to prove the claim when s = 0, because otherwise one
may define a new sequence by c̃0 = c0 − s, and c̃n = cn for n > 0. Then∑
cnr
n −
∑
n c̃nr
n = s for all r, and it is clear that
∑
c̃n = 0.
Now by telescoping, one has the identity
(1− r)
N∑
n=0
snr
n + sNr
N+1 =
N∑
0
snr
n −
N∑
0
snr
n+1 + sNr
N+1
=
N∑
0
(sn − sn−1)rn − sNrN+1 + sNrN+1
=
N∑
0
cnr
n.
We can let N →∞ to obtain
∞∑
0
cnr
n = (1− r)
∞∑
0
snr
n, (1)
whenever |r| < 1. Now assuming s = 0, we fix an � > 0, and we choose some M
so that n ≥M implies |sn| < �. Then we can write
∞∑
0
cnr
n = (1− r)
M∑
0
snr
n + (1− r)
∑
n>M
snr
n.
The first term on the right side clearly aproaches 0 as r → 1, and for the second
one, we have ∣∣∣∣(1− r) ∑
n>M
snr
n
∣∣∣∣ ≤ (1− r) ∑
n>M
�rn = �rM+1.
From the last expression, it is clear that
lim sup
r→1−
∣∣∣∣∑ cnrn
∣∣∣∣ ≤ �,
which proves the claim since � was arbitrary.
Part b: If cn = (−1)n, then sn alternates between 0 and 1 depending on whether
n is even or odd. So clearly sn does not converge. However
∑
(−1)nrn = 1
1+r
,
which clearly approaches 1/2 as r → 1 from the left.
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Part c: As before, we may assume that σ = 0. Replacing cn with sn in equation
(1) it holds that
∞∑
0
snr
n = (1− r)
∞∑
0
nσnr
n,
for all |r| < 1. Hence ∞∑ 0 cnr n = (1− r) ∞∑ 0 snr n = (1− r)2 ∞∑ 0 nσnr n. Now we fix � > 0, and choose M so that n ≥M implies |σn| < �. Then ∞∑ 0 cnr n = (1− r)2 ∑ n≤M nσnr n + (1− r)2 ∑ n>M
nσnr
n.
The first term on the right side clearly approaches 0 as r → 1, and for the
second one, we have∣∣∣∣(1− r)2 ∑
n>M
nσnr
n
∣∣∣∣ ≤ (1− r)2 ∑
n>M
n�rn ≤ �r,
where we used
∑
n>M nr
n ≤
∑
n≥0 nr
n = r(1− r)2. From the last expression,
it is clear that
lim sup
r→1−
∣∣∣∣∑ cnrn
∣∣∣∣ ≤ �,
which proves the claim since � was arbitrary.
Part d: We first show that if cn is Cesaro summable to σ, then cn/n→ 0. Indeed,
let σn =
1
n
∑n
1 sk. Then σn → σ, hence
cn
n
= σn −
(n−1)
n
σn−1 → σ − 1 · σ = 0,
as desired. Therefore if cn = n(−1)n then it cannot be Cesaro summable, since
cn/n = (−1)n. However
∞∑
n=0
n(−1)nrn = r
d
dr
[
1
1 + r
]
= −
r
(1 + r)2
,
which clearly approaches −1/4 as r → 1 from the left.
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Chapter 2, Exercise 15
Letting ω = eix and summing the geometric series, we have
NFN (x) =
N−1∑
n=0
ω−n − ωn+1
1− ω
=
1
1− ω
[N−1∑
n=0
ω−n −
N−1∑
k=0
ωn+1
]
=
1
1− ω
[
1− ω−N
1− ω−1
−
ω(1− ωN )
1− ω
]
=
1
1− ω
[
(1− ω−N )(1− ω)− ω(1− ωN )(1− ω−1)
(1− ω−1)(1− ω)
]
=
1
1− ω
[
ω1−N + ωN+1 − 2ω + 2− ω−N − ωN
2− ω − ω−1
]
=
1
1− ω
[
(2− ωN − ω−N )(1− ω)
2− ω − ω−1
]
=
2− ωN − ω−N
2− ω − ω−1
=
−(ωN/2 − ω−N/2)2
−(ω1/2 − ω−1/2)2
,
which completes the proof, after noting that ωk/2 − ω−k/2 = eikx/2 − e−ikx/2 =
2i sin(kx/2).
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Chapter 3, Exercise 19
Note that∫ x
0
DN (t)dt =
∫ x
0
sin((N + 1/2)t)
[
1
sin(t/2)
−
2
t
]
dt+
∫ x
0
sin((N + 1/2)t)
t/2
dt.
Let’s call the terms on the right side as A and B, respectively. Then
A ≤
∫ x
0
| sin((N + 1/2)t)|
∣∣∣∣ 1sin(t/2) − 2t
∣∣∣∣dt
≤
∫ π
0
1 ·
∣∣∣∣ 1sin(t/2) − 2t
∣∣∣∣dt,
which proves the desired bound on A. Here we are using the fact that | sin((N+
1/2)t)| ≤ 1, that x ≤ π, and that the function 1
sin(t/2)
− 2
t
extends continuously
to [−π, π].
Now for B, we have by substitution u = (N + 1/2)t:
B =
∫ x
0
sin((N + 1/2)t)
t/2
dt
= 2
∫ (N+1/2)x
0
sinu
u
du
which is of course uniformly bounded in N and x by 2 supa>0
∫ a
0
sinu
u
du, which
in turn is finite by the result of Exercise 12 of the same chapter.
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Chapter 3, Exercise 20
Recall that there was a typo in the problem: there should be a “+O(N−1)
as N →∞” on the right-hand side of the last expression.
Setting x = π/N and using the same chain of equalities as in the previous
exercise, we have∫ π/N
0
DN (t)dt =
∫ π/N
0
sin((N + 1/2)t)
[
1
sin(t/2)
−
2
t
]
dt+ 2
∫ π(1+ 1
2N
)
0
sinu
u
du.
Since | sin((N + 1/2)t)| ≤ 1 and since 1
sin(t/2)
− 2
t
is bounded on [0, π], the first
term on the RHS is bounded by Cπ/N = O(N−1). Similarly
∫ π(1+ 1
2N
)
0
sinu
u
du =
∫ π
0
sinu
u
du+
∫ π(1+ 1
2N
)
π
sinu
u
du =
∫ π
0
sinu
u
du+O(N−1).
This shows that
1
2
∫ π/N
0
(DN (t)− 1)dt =
∫ π
0
sinu
u
du+O(N−1),
thus giving a lower bound for the desired maximum. To prove an upper bound,
one may replace π/N with any sequence (xn) such that xn ∈ [0, π/n], then go
through this same chain of equalities and note that it is still bounded above by∫ π
0
sinu
u
du+O(N−1).
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