Partiel - Normalesup.org

un=n!
nnvn=(−1)n (−1)nn+ ln nwn= sin (−1)n n
xn=(ln n)α
nα∈Ryn= (−1)n(ln n)α nα∈R
un+1
un=(n+ 1)! nn n!(n+ 1)n+1 =nn (n+ 1)n=1 + 1 n+ 1n
=enln(1+1/(n+1)) →1/e
1/e < 1
vn=(−1)n
(−1)nn+ ln n=1 n
1
1 + ln n
(−1)nn∼1
n
1/n vn
wn= (−1)nsin(1/n)an= sin(1/n) sin
[0, π/2] sin 0 = 0
wn
an=(ln n)α
nf(t) = (ln t)α tt>CαCα
α ann
ynα∈R
Z+∞
0
sin t
et−1dt Z1
−1
dt
1−√1−t2Z+∞
1
ln(1 + 1/t)
(t2−1)αdt (α∈R)

0et= 1 + t+o(t) limt→0sin t
et−1= limt→0t
t= 1
t7→ sin t
et−10
+∞sin t
et−1≤1 et−1t∈R+1 et−1∼1 et+∞
+∞t7→ sin t
et−1
t7→ sin t
et−1[0,+∞[
t7→ 1/(1 −√1−t2)
Z1
−1
dt
1−√1−t2= 2 Z1
0
dt
1−√1−t2
0
1
1−√1−t2=1
1−(1 −1
2t2+o(t2)) ∼1
t2
t7→ 1/t20
1t= 1 + u((t−1)(t+ 1))α
ln(1 + 1/(1 + u))
uα(u+ 2)α∼ln 2 2αuα
u7→ 1/uα0α < 1
α1α < 1
+∞(t2−1)α∼t2αln(1 + 1/t)∼1/t
ln(1+1/t)
(t2−1)α∼1/t2α+1 +∞
2α+ 1 >1α > 0
t7→ ln(1+1/t)
(t2−1)α[1,+∞[α∈]0,1[
(fn)n∈N[0,1]
∀t∈[0,1], fn(t) = (1 −t2)n.
(fn)f
[0,1]
[ε, 1] ε∈]0,1[
lim
n→+∞Z1
0
(1 −t2)ndt = 0.
(gn)
∀x∈[−1,1], gn(x) = Rx
0(1 −t2)ndt
R1
0(1 −t2)ndt .
gn

∀n∈N,1
2(n+ 1) ≤Z1
0
(1 −t2)ndt
t(1 −t2)n≤(1 −t2)n[0,1]
∀n∈N,∀x∈[−1,1],|gn(x)−1| ≤ 2(n+ 1)(1 −x2)n
(gn)
(gn) [ε, 1] ε∈]0,1[
n∈Nx∈[−1,1] hn(x) = Rx
0gn(t)dt (hn)
[−1,1] x7→ |x|
0fn(0) = 1 n∈Nlimn→+∞fn(0) = 1
t∈]0,1] |1−t2|<1 limn→+∞fn(t) = 0
(fn)f[0,1]
f(t)=1 t= 0
f(t)=0 t∈]0,1]
(fn)
f0
[0,1]
ε∈]0,1[
n∈N∗fn[0,1]
sup
[ε,1] |fn|= (1 −ε2)n
|(1 −ε2)|<1 limn→+∞sup[ε,1] |fn|= 0
(fn) 0 = f|[ε,1] [ε, 1]
ε∈]0,1[
Z1
0
(1 −t2)ndt =Zε
0
fn(t)dt
| {z }
+Z1
ε
fn(t)dt
| {z }
I1I2
I1t∈[0, ε]|fn(t)| ≤ 1|I1| ≤ ε
I2(fn) 0
[ε, 1]
lim
n→+∞Z1
ε
fn(t)dt =Z1
ε
lim
n→+∞fn(t)dt =Z1
ε
0dt = 0
N∈Nn≥N|I2| ≤ ε
N∈Nn≥N
Z1
0
(1 −t2)ndt≤ |I1|+|I2| ≤ 2ε
ε∈]0,1[ limn→+∞R1
0(1 −t2)ndt = 0
t7→ (1−t2)ngn

n∈Nt∈[0,1] t(1 −t2)n≤(1 −t2)n
[0,1] Z1
0
t(1 −t2)ndt ≤Z1
0
(1 −t2)ndt
Z1
0
t(1 −t2)n=(1 −t2)n
2(n+ 1) 1
0
=1
2(n+ 1)
1
2(n+ 1) ≤Z1
0
(1 −t2)ndt
n∈Nx∈[−1,1]
|gn(x)−1|=Rx
0(1 −t2)ndt
R1
0(1 −t2)ndt −1
=Rx
0(1 −t2)ndt −R1 0(1 −t2)ndt
R1
0(1 −t2)ndt 
=−R1
x(1 −t2)ndt
R1
0(1 −t2)ndt
≤2(n+ 1) Z1
x
(1 −t2)ndt
≤2(n+ 1)(1 −x) sup
[x,1]
fn= 2(n+ 1)(1 −x)(1 −x2)n
≤2(n+ 1)(1 −x2)n
∀n∈N,∀x∈[−1,1],|gn(x)−1| ≤ 2(n+ 1)(1 −x2)n
x= 0 gn(0) = 0 nlimn→+∞gn(0) = 0
x > 0|1−x2|<1 limn→+∞2(n+ 1)(1 −x2)n= 0
limn→+∞|gn(x)−1|= 0
gn(gn)
g[−1,1]
g(t) = −1t∈[−1,0[
g(t)=0 t= 0
g(t)=1 t∈]0,1]
ε∈]0,1[
n∈N
sup
[ε,1] |gn−1| ≤ 2(n+ 1) sup x∈[ε,1]
(1 −x2)n= 2(n+ 1)(1 −ε2)n
|1−ε2|<1 limn→+∞(sup[ε,1] |gn−1|) = 0
(gn) [ε, 1]
hn
[0,1] [−1,1]
hn
n∈Nsupx∈[0,1] |hn(x)−x|(1 −t2)n≥0t∈
[0,1] gn(x)≤1x∈[0,1] x7→ |hn(x)−x|
x= 1
sup
x∈[0,1] |hn(x)−x|= 1 −hn(1) = Z1
0
1−gn(t)dt

lim+∞(hn(1) −1) = 0
(hn) [−1,1] x7→ |x|