PC* Corrigé DM 3 Exercice 1 Intégrale de Wallis et formule de Stirling

2016 −2017
t7→ sinn(t) [0, π/2]
∀n∈N, In>0
f[a, b]R+Zb
a
f(u)du = 0
∀t∈[a, b], f(t) = 0
∀n∈N, In+1 −In=Zπ/2
0
sinn(t)(sin t−1)dt
[0, π/2] ∀n∈N, In+1 ≤In
I0=Zπ/2
0
1dt =π
2I1= [−cos t]π/2
0= 1
n n ≥2
In=Zπ/2
0
cosn−1(t) cos(t)dt =cosn−1(t) sin(t)π/2
0−Zπ/2
0
(n−1) cosn−2(t)(−sin t) sin t(t)dt
In= 0 + (n−1) Zπ/2
0
cosn−2(t)(1 −cos2t)dt = (n−1)In−2−(n−1)In
∀ ∈ N, n ≥2, In=n−1
nIn−2
n n N∗InIn−1=π
2n(Pn)
n= 0 I0I1=π
2×1 = π 2×1
(Pn)n
nN∗In+1 =n
n+ 1In−1
In+1In=n
n+ 1In−1In=n n+ 1 ×π 2n
In+1In=π
2(n+ 1) (Pn+1)
n
nN∗In+1 ≤In≤In−1
In−1
n
n+ 1In−1≤In≤In−1
In>0n
n+ 1InIn−1≤I2
n≤InIn−1
π
2(n+ 1) ≤I2
n≤π
2n
In>0rn
n+ 1 ≤In√2n
√π≤1 lim
n→+∞
In√2n
√π= 1
In∼√π
√2n
k∈NI2k+2
I2k
=2k+ 1
2k+ 2
p−1
Y
k=0
I2k+2
I2k
=I2p
I0
=
p−1
Y
k=0
2k+ 1
2k+ 2 =1×3× ··· × (2p−1) 2×4× ··· × (2p)=(2p)! (2 ×4× ··· × (2p))2

2016 −2017
I2p=(2p)!
(2p×p!)2I0=(2p)! 22p(p!)2
π
2
p I2p+1I2p=π
2(2p+ 1)
I2pI2p+1 =π
2(2p+ 1)
22p(p!)2
(2p)!
2
πI2p+1 =22p(p!)2 (2p+ 1)!
K n!∼K nne−n√n
I2p∼(2p)2pe−2p√2pK
22p(ppe−p√pK)2
π
2=π
K√2pI2p∼√π
√2×2p
lim
p→+∞
I2p
I2p
= 1 = lim
p→+∞
π
K√2p
2√p
√π=√2π
KK=√2π
n!∼nne−n√2πn
X(−1)n
2n+ 1
nNun=1
2n+ 1 lim un= 0 un+1 =1 2n+ 3 ≤un
(un)X(−1)nun
k(−1)k
2k+ 1 =Z1
0
(−1)kt2kdt =Z1
0
(−t2)kdt
n
X
k=0
uk=
n
X
k=0 Z1
0
(−t2)kdt =Z1
0 n
X
k=0
(−t2)k!dt
∀t∈[0,1],−t26= 1
Sn=
n
X
k=0
uk=Z1
0
1−(−t2)n+1
1−(−t2)dt =Z1
0
1
1 + t2dt −Z1
0
(−t2)n+1
1 + t2dt
n Kn=Z1
0
(−t2)n+1
1 + t2dt
|Kn| ≤ Z1
0
t2n+2
1 + t2dt ≤Z1
0
t2n+2dt =1
2n+ 3 lim
n→+∞Kn= 0
n Sn= [arctan(t)]1
0−Kn=π
4−Kn
(Sn)π/4
X(−1)n
2n+ 1 S=π/4
S=
+∞
X
n=0
(−1)n
2n+ 1 =π 4∀n∈N, Rn=S−Sn=Kn=Z1
0
(−t2)n+1
1 + t2dt
X π
4−
n
X
k=0
(−1)k
2k+ 1!=XRn=XZ1
0
(−t2)n+1
1 + t2dt

2016 −2017
n
Tn=Z1
0 n
X
k=0
(−t2)k+1!dt
1 + t2=−Z1
0 n
X
k=0
(−t2)k!t2
1 + t2dt
Tn=−Z1
0
1−(−t2)n+1
1 + t2
t2
1 + t2dt =−Z1
0
t2dt
(1 + t2)2+K0
n
|K0
n| ≤ Z1
0
t2n+4dt =1
2n+ 5
lim
n→+∞Tn=−Z1
0
t2
(1 + t2)2dt
π
4=Z1
0
dt
1 + t2=t1 1 + t21
0−Z1
0
t×−2t
(1 + t2)2dt
Z1
0
t2
(1 + t2)2dt =π 8−1 4
+∞
X
n=0
Rn=1
4−π 8
a > 0X(−1)n
an + 1
Sn(a) =
n
X
k=0 Z1
0
(−ta)kdt =Z1
0
1−(−ta)n+1
1 + tadt =Z1
0
dt
1 + ta−Kn(a)
|Kn(a)| ≤ Z1
0
tna+adt =1
(n+ 1)a+ 1 lim
n→+∞Kn(a)=0
S(a)S(a)−
Sn(a)Kn(a)
S(a) = Z1
0
dt
1 + tadt
Rn(a) = Kn(a) = Z1
0
(−ta)n+1
1 + tadt
n
X
k=0
Kk(a) = Z1
0 n
X
k=0
(−ta)k!−ta
1 + tadt =Z1
0
−ta
(1 + ta)2+Z1
0
(−ta)n+1
(1 + ta)2tadt
Z1
0
t(n+2)adt =1
(n+ 2)a+ 1
+∞
X
n=0
Rn(a) = −Z1
0
ta
(1 + ta)2dt
+∞
X
n=0
(−1)n
n+ 1 = 1−1 2+1 3−1 4+··· =Z1
0
dt
1 + t= ln 2
Rn(1) =
+∞
X
k=n+1
(−1)k
k+ 1
+∞
X
n=0
Rn(1) = −Z1
0
t
(1 + t)2dt =−ln 2 + 1 2
Z Q
∀(p, q)∈Z2, p + (−q) = p−q∈Z(Z,+) (R,+)

2016 −2017
a b Qp, q, p0, q0Zq > 0
q0>0a=p/q b =p0/q0a+ (−b) = a−b=pq0−p0q
qq0∈Q
(Z,+) (Q,+) (R,+)
G(R,+) g∈G0 = g+(−g)∈G−g= 0+(−g)∈G
2g=g+ (−(−g)) ∈G∀n∈N, ng ∈G
(n+ 1)g=ng + (−(−g))
∀n∈N,−ng =n(−g)∈G−g∈G∀p∈Z, pg ∈G
a∈RaZ={xa / x ∈Z}aZa x, y
p, q Z
x=pa y =qa x + (−y) = (p−q)a∈aZ
(aZ,+) (R,+)
G(R,+) a∈G
aZ⊂G
G(R,+) {0}
G+G
G α −α
G G+
m= inf(A)⇔ ∀a∈A, m ≤a∀m0> m, m0A
m= inf(A)⇔ ∀a∈A, m ≤a∀m0> m, ∃a0∈A / m ≤a0< m0
A=]0,1[ inf(A) = 0 0 6∈ A
G+
G=πZG+={nπ , n ∈N∗}inf G+=π
α= inf Q+Q+0≤α
∀n∈N∗,1/n ∈Q+0≤α≤1/n α = 0
G(R,+) α= inf G+
α > 0α < 3α/2 3α/2G+
g G+α≤g < 3α
2
α < g g
G+g0∈G+
α≤g0< g
α≤g0< g < 3α/2d(g, g0) = g−g0< d(3α/2, α/2) = α/2
g−g0=g+ (−g0)∈G g −g0>0g−g0∈G+
g−g0≥α
α=g∈G+
α∈G αZ⊂G

2016 −2017
x∈G p =bx
αc
p≤x
α< p + 1 pα ≤x < (p+ 1)α0≤x−pα < α
α x −pα 6∈ G+
G G
x−pα = 0 x=pα G ⊂αZ
G=αZ
G(R,+) {0}
inf G+= 0
nN1/(n+ 1) G+
gn∈G+0< gn<1
n+ 1
x
nNyn=x
gngn
pg p ∈Zg∈G G
x
gn≤x gn
<x
gn+ 1
gn>0yn≤x<yn+gn0≤x−yn< gn<1
n+ 1
(yn)n∈N=x
gngnn∈N
G
x
G
r G ={n+kr/(n, k)∈Z2}
rQr=p0
q0
p0∈Z
q0∈N∗p0q0±1
G=n+kp0
q0
(n, k)∈Z2=nq0+kp0
q0
,(n, k)∈Z
nq0+kp0
p0q0G=nN
q0, N ∈Zo
inf(G+) = 1
q0
α= inf G+
G=αZ
1 = 1 + 0 ×r∈G q 1 = qα
r= 0 + 1 ×r∈G p r =pα
r=p
q∈Q
G αZr∈Q
GRr6∈ Q