Corrections d'exercices : Dénombrabilité, Séries

ANϕ
n∈NA=ϕ(n)n
A
n∈A n /∈ϕ(n)A=ϕ(n)
n /∈A n /∈ϕ(n)n∈A
n
anxn+· · · +a1x+a0= 0 a0, a1, . . . , an∈Zan6= 0
Z∗×Zn
n
+∞
X
n=0
1
2n+1 =1 2
1
1−1/2
n
X
k=0
1
2k+1 <
+∞
X
k=0
1
2k+1 = 1
[0 ; 1]
(xn)
[0 ; 1]
[0 ; 1] (un)n∈N
(xn)n∈N`
[0 ; 1] N∈NuN=`
(xn)n∈N`
[`−1/2N+2 ;`+ 1/2N+2] = [uN−1/2N+2 ;uN+ 1/2N+2]
n≥N xn
ENEn
J0 ; nKN
E=[
n∈N
En
EnE
i1, . . . , ikAN
n(A)=2i1+· · · + 2ik
2
1
(p+q)2≤1 p2+q2≤2 (p+q)2
1
(p+q)2α(p,q)∈N∗2

In=(p, q)∈N∗2p+q=n
n−1
n2αn≥2
α > 1
|z|<1
1
1−z2n+1 =
+∞
X
k=0
z2n+1k
+∞
X
n=0
z2n
1−z2n+1 =
+∞
X
n=0
z2n
+∞
X
k=0
z2n+1k=
+∞
X
n=0
+∞
X
k=0
z2n(2k+1)
p
p= 2n(2k+ 1) n, k ∈N
N∗
An=2n(2k+ 1) k∈N
(zp)p∈N∗
+∞
X
p=1
zp=
+∞
X
n=0 X
m∈An
zm=
+∞
X
n=0
+∞
X
k=0
z2n(2k+1)
+∞
X
n=0
z2n
1−z2n+1 =
+∞
X
p=1
zp=z
1−z
v∈`1(Z)vn−−−−−−→
|n|→+∞0 (vn)M
|ukvn−k| ≤ M|uk|(ukvn−k)k∈Z
k∈Z(|ukvn−k|)n∈Z
X
n∈Z
|ukvn−k|=|uk|X
n∈Z
|vn−k|=|uk|X
n∈Z
|vn|
|uk|Pn∈Z|vn|!k∈Z
(ukvn−k)(n,k)∈Z2
X
(n,k)∈Z2
|ukvn−k|=X
n∈ZX
k∈Z
|uk||vn−k|<+∞
X
k∈Z
ukvn−k ≤X
k∈Z
|uk||vn−k|
u∗v∈`1(Z)
X
(n,k)∈Z2
ukvn−k=X
n∈ZX
k∈Z
ukvn−k=X
k∈ZX
n∈Z
ukvn−k
X
n∈Z
(u∗v)n=X
k∈Z
ukX
n∈Z
vn−k=X
k∈Z
ukX
`∈Z
v`
(u∗v)n=X
k+`=n
ukv`= (v∗u)n
((u∗v)∗w)n=X
k+`+m=n
ukv`wm= (u∗(v∗w))n
ε εn=δn,0u∗ε=u ε
u un=δ0,n −δ1,n
u v u ∗v=ε
vn−vn−1=εn=δ0,n
n∈N, vn=v0vn−−−−−→
n→+∞0
n∈N, vn= 0 n < 0, vn= 0
n= 0 vn−vn−1=δ0,n 0=1
u(`1(Z),∗)

|q||n|n∈Z
Z=N∗∪ {0} ∪ Z∗
−
|q||n|n∈N∗P|q|n
|q||n|n∈Z∗
−
|q||n|n∈Z
X
n∈Z
q|n|=X
n∈N∗
qn+1+ X
n∈Z∗
−
q−n= 1 + 2
+∞
X
n=1
qn=1 + q
1−q
r|n|einθ n∈Z=r|n|n∈Z
Z=N∗∪ {0} ∪ Z∗
−
r|n|n∈N∗Prn
r|n|n∈Z∗
−
r|n|n∈Z
X
n∈Z
r|n|einθ =X
n∈N∗
rneinθ+1+ X
n∈Z∗
−
r−neinθ = 1+ reiθ
1−reiθ+re−iθ 1−re−iθ=1−r2 1−2rcos θ+r2
n
X
k=0
|vk| ≤
+∞
X
n=0
|un|<+∞
Pn≥0vn
n∈N
p(n) = maxσ−1(k)0≤k≤n
ε > 0N∈NPn≥N+1|un| ≤ ε
M≥p(N)
M
X
n=0
vn−
N
X
n=0
un ≤X
n≥N+1
|un| ≤ ε
M
X
n=0
vn−
+∞
X
n=0
un ≤2ε
+∞
X
n=0
vn=
+∞
X
n=0
un
P1
n21 n2n≥1
1
σ(n)2n≥1
Pn≥11
σ(n)2
1
nn≥1
Pn≥11
σ(n)
Sn=
n
X
k=1
σ(k)
k2
S2n−Sn=
2n
X
k=n+1
σ(k)
k2≥1 4n2
2n
X
k=n+1
σ(k)
σ(n+ 1), . . . , σ(2n) 1, . . . , n
S2n−Sn≥1
4n2
n
X
k=1
k=n+ 1
8n≥1 8
(Sn)
Sn=
n
X
k=2
σ(k)
k2ln k

S2n+1 −S2n≥1
22(n+1) ln 2n+1
2n+1
X
k=2n+1
σ(k)≥1
22(n+1) ln 2n+1
2n
X
k=1
k
σ(k)k
S2n+1 −S2n≥2n(2n+ 1)
22n+3(n+ 1) ln 2 ∼1 8 ln 2
1
n
P1/n
PS2n+1 −S2n(S2n) +∞
+∞
X
n=0
v2
n=
+∞
X
n=0
u2
n
ab ≤1
2a2+b2
|unvn| ≤ 1
2u2
n+v2 n
P|unvn|
+∞
X
n=0
|unvn| ≤ 1
2
+∞
X
n=0
u2
n+1
2
+∞
X
n=0
v2
n=
+∞
X
n=0
u2
n
σ= IdN
sup(+∞
X
n=0
|unvn|σN)=
+∞
X
n=0
u2
n
+∞
X
n=0
|unvn| ≥ 0
ε > 0
Pu2
nN∈N
+∞
X
n=N
u2
n≤ε
(un)M > 0
N0> N
∀n≥N0,|un| ≤ ε
M(N+ 1)
σN
σ(n) = 
N0+n n ∈ {0, . . . , N}
n−N0n∈ {N0, . . . , N0+N}
n
+∞
X
n=0
|unvn| ≤
N−1
X
n=0
|un|ε
M(N+ 1) +
N0+N−1
X
n=N0
ε
M(N+ 1)|un−N0|+ε≤3ε
inf(+∞
X
n=0
|unvn|σN)= 0
N∈NAN=n∈N,|zn| ≤ N
n, m ∈ANznzm1/2
n AN
N+ 1/2
Card AN×π×1
22
≤πN+1
22

Card AN≤(2N+ 1)2
(|zn|)
|z(2N+1)2+1|> N
1
|zp|3= O1
p3/2
1/z3
n
x7→ 1
xα
Z+∞
n+1
dx
xα≤
+∞
X
k=n+1
1
kα≤Z+∞
n
dx
xα
+∞
X
k=n+1
1
kα∼1 α−1
1
nα−1
P+∞
n=0 P+∞ k=n+1 1
kαα > 2
uk,n =1
kαk > n uk,n = 0
n≥1Pk≥0|uk,n|Pn≥0P+∞
k=0|uk,n|
+∞
X
n=0
+∞
X
k=0
uk,n =
+∞
X
k=0
+∞
X
n=0
uk,n
+∞
X
n=0
uk,n =
k−1
X
n=0
1
kα=1 kα−1
+∞
X
n=0
+∞
X
k=n+1
1
kα=
+∞
X
k=1
1
kα−1
Pp≥1up,q
+∞
X
p=1
|up,q|=|a|2q−1
1− |a|2q−1
|a|2q−1
1−|a|2q−1
(up,q)p,q≥1
+∞
X
q=1
+∞
X
p=1
up,q =
+∞
X
p=1
+∞
X
q=1
up,q
+∞
X
q=1
a2q−1
1−a2q−1=
+∞
X
p=1
ap
1−a2p
P+∞
p=0 ap,q = 0 P+∞ q=0 P+∞ p=0 ap,q = 0
P+∞
q=0 ap,q =1
p+1 −1 p+2 P+∞
p=0 P+∞ q=0 ap,q = 1
(ap,q)(p,q)∈N2
q∈N∗Pp≥01
(p+q2)(p+q2+1)
1
(p+q2)(p+q2+1) ∼1 p2
+∞
X
p=0
1
(p+q2)(p+q2+ 1) =
+∞
X
p=01
p+q2−1 p+q2+ 1=1 q2
Pq≥1P+∞
p=0 1
(p+q2)(p+q2+1) =Pq≥11 q2
1
(p+q2)(p+q2+ 1)(p,q)∈N×N∗