Idée de corrigé

a < 1
(un)
t7−→ 1
2(1 + t2)
[0,1]
0≤x≤a < 1 =⇒f(0) ≤f(x)≤f(a)< f(1).
0≤x≤a < 1 = ⇒1
2≤f(x)≤f(a)<1.
n∈Nx7−→ xn
[0,1]
0≤(f(t))n≤(f(a))n∀t∈[0, a]
a0< a f
Za
0
0 dt≤Za
0
(f(t))ndt≤Za
0
(f(a))ndt
0≤un≤a(f(a))n∀n∈N
0< a < 10< f(a)<1n7−→ (f(a))n
(un)n
+∞
Pun
n un+1 =Sn+1 −Sn(Sn)
n+ 1
r
n
X
k=0
rk!×(1 −r) =
n
X
k=0
rk−
n+1
X
k=1
rk
= 1 −rn+1
(1 −r)6= 0
n
X
k=0
rk=1−rn+1
1−r
0≤t≤a0≤f(t)<1
n
X
k=0
(f(t))k=1−(f(t))n+1
1−f(t)∀t∈[0, a]
1−f(t)
[0, a]
Za
0n
X
k=0
(f(t))k!dt=Za
0
1−(f(t))n+1
1−f(t)dt
n
X
k=0 Za
0
(f(t))kdt=Za
0
1−(f(t))n+1
1−f(t)dt
n
X
k=0
uk=Za
0
1−(f(t))n+1
1−f(t)dt Sn=Za
0
1−(f(t))n+1
1−f(t)dt

f
Sn=Za
0
1−(f(t))n+1
1
2(1 −t2)dt
Sn= 2 Za
0
1−(f(t))n+1
1−t2dt
0< a 0<1−t20≤(f(t))n+1 <1∀t∈[0, a]
Sn≤2Za
0
dt
1−t2
(Sn)
Pun
0≤(f(t))n+1 ≤(f(a))n+1 ∀t∈[0, a]
0<1−a2≤1−t2∀t∈[0, a]
t7−→ 1−t2[0, a]
0≤(f(t))n+1
1−t2≤(f(a))n+1 1−a2∀t∈[0, a]
a0< a
0≤Za
0
(f(t))n+1
1−t2dt≤a(f(a))n+1 1−a2
n7−→ Za
0
(f(t))n+1
1−t2dt
n+∞
Sn= 2 Za
0
1−(f(t))n+1
1−t2dt
Sn=Za
0
2
1−t2dt−2Za
0
(f(t))n+1
1−t2dt
n+∞
lim
n→+∞
Sn=Za
0
2
1−t2dt
I=Za
0
2
1−t2dt
2
1−t2
2
1−t2=1 1 + t+1 1−t
1−t1 + t
Za
0
2
1−t2dt= ln(1 + a)−ln(1 −a)
I= ln 1 + a
1−a

a= 1.
(un)
0≤t≤10≤t2≤t0≤1 + t2≤1 + t
0≤(1 + t2)n≤(1 + t)n∀t∈[0,1]
0≤(1 + t2)n
2n≤(1 + t)n 2n∀t∈[0,1]
[0,1]
0<1
Z1
0
0 dt≤Z1
0
(1 + t2)n
2ndt≤Z1
0
(1 + t)n
2ndt
0≤un≤In
In=1
2nZ1
0
(1 + t)ndt
x7−→ t=x−1C1R
Z1
0
(1 + t)ndt=Z2
1
xndx
Z1
0
(1 + t)ndt="xn+1
n+ 1#2
1
=2n+1 −1
n+ 1
In=2
n+ 1 −1 2n(n+ 1)
Inn+∞
un+∞
P(−1)nun
n
X
k=0
(−1)k(f(t))k
t6= 1 n
r=−f(t)
n
X
k=0
(−1)k(f(t))k=1+(−1)n(f(t))n+1
1 + f(t)
= 2 1 + (−1)n(f(t))n+1
3 + t2
t= 1
n
X
k=0
(−1)k(f(1))k= 0 nimpair
= 1 npair

[0,1]
n
X
k=0
(−1)kuk=
n
X
k=0
(−1)kZ1
0
(f(t))kdt
=Z1
0
n
X
k=0
(−1)k(f(t))kdt
= 2 Z1
0
1 + (−1)n(f(t))n+1
3 + t2dt
f0≤f(t)≤10≤(f(t))n≤1∀t∈
[0,1],∀n∈N
0≤1
3 + t2≤1∀t∈[0,1]
0≤(f(t))n+1
3 + t2≤(f(t))n+1 ∀t∈[0,1]
[0,1]
[0,1]
Z1
0
0 dt≤Z1
0
(f(t))n+1
3 + t2dt≤Z1
0
(f(t))n+1 dt
0≤Jn≤un+1
n+∞Jn
lim
n→+∞
Jn= 0
Jn
n
X
k=0
(−1)kuk=Z1
0
2
3 + t2dt+ 2 (−1)nJn
n+∞
lim
n→+∞
n
X
k=0
(−1)kuk=Z1
0
2
3 + t2dt
x7−→ t=√3x
"0,√3
3#[0,1]
Z√3
3
0
2
3+3x2√3 dx=Z1
0
2
3 + t2dt
u7−→ x= tan(u)0,π
6
"0,√3
3#
Zπ
6
0
1
1 + tan2(u)(1 + tan2(u)) du=Z√3
3
0
1
1 + x2dx

Z1
0
2
3 + t2dt=2√3 3×Zπ
6
0
1
1 + tan2(u)(1 + tan2(u)) du
=π√3
9
(−1)nun
+∞
X
k=0
(−1)kuk=π√3
9.
Pun
[0,1]
0≤t2≤1 =⇒0≤t2≤2t2≤1 + t2
0≤t2≤1 + t2
20≤t2≤f(t)∀t∈[0,1]
0≤t2n≤(f(t))n∀t∈[0,1]
[0,1]
Z1
0
t2ndt≤Z1
0
(f(t))ndt
1
2n+ 1 ≤un∀n∈N
n2n+ 1 ≤2 (n+ 1)
1
2 (n+ 1) ≤1 2n+ 1
1
2
n
X
k=0
1
(n+ 1) ≤
n
X
k=0
1
2n+ 1
+∞n
Sn+∞+∞
Pun+∞
X(un)2Xun
n+ 1
0≤un≤InIn=2
n+ 1 −1 2n(n+ 1) ≤2 n+ 1
0≤u2
n≤I2
nI2 n≤4
(n+ 1)2n∈N
X(un)2
0≤un
n+ 1 ≤In n+ 1
In
n+ 1 ≤2 (n+ 1)2n∈N
Xun
n+ 1