Suites numériques : Exercices MPSI2 - Lycée Champollion

no14
(un),(vn)l1, l2l1< l2
un< vnl1⩽l2
(un) (vn)
(un) (vn) [0; 1] lim unvn= 1
n∈Nun⩽a vn⩽b a, b ∈Rlim un+vn=a+b
(un+vn) (un−vn)a, b ∈R
lim (u2
n+unvn+v2 n)=0
ε
ε
(ln(n)) +∞
(e−n) 0
(1
n) 0
(√n) +∞
1−sin2(2n)
√n
√nj1
√nk
√n
n
√n2+n+ 1 −√n2−n+ 1
an−bn
an+bna, b ∈R∗
+
sin 1
n1/n
Sn
Pn
k=1 √k
Pn
k=1
1
√k
Pn
k=1
1
n2+k2
P2n
k=n+1
1
k2
Pn
k=1
n
n2+k
Pn
k=1
1
√n2+k
(un)un+1
unl∈R
l∈R+l= +∞
l < 1 (un) 0
l > 1 (un) +∞
l= 1 (un)
(un)n
√unl∈R
(un)

|un|+∞(un) +∞
−∞
un>0 (nun) +∞
un>1 (un
n) +∞
(u4
n) (u2 n)
unun01
un
+∞
unun+1
un1
(un)
(u2n) (u2n+1) (un)
(un)∀n∈N∗, un=1
nPn
k=1
1
√k
(un) [0,1] l∈[0; 1]
n∈N∗un⩽un
2+1
2√nl
n∈Nxn= cos(x) [0; 1]
xn
(xn)
n∈N∗xn+xn−1+··· +x2+x= 1 R+
xn
n∈N∗xn∈[1/2; 1]
(xn)l[1/2; 1[
l
α∈R\πZ(un) = (cos(nα)) (vn) = (sin(nα))
n∈Nun+1 vn+1 unvn
(un) (vn)
(un) (vn)
α∈πZ
a, b ∈R+
2√ab ⩽a+b
(un),(vn)
u0=a, v0=b, ∀n∈N, un+1 =√unvn, vn+1 =un+vn
2.
(un) (vn)M(a, b)
a b

M(a, a)M(a, 0) M(λa, λb)M(a, b)λ∈R+
(un) 0 n∈NSn=Pn
k=0(−1)kuk
(Sn)l n ∈N|Sn−l|⩽un+1
n∈N∗un=Pn
k=1
1
√k−2√n vn=Pn
k=1
1
√k−2√n+ 1
(un) (vn)Pn
k=1
1
√k
1
√n·Pn
k=1
1
√k
(eiln(n))
(un),(vn)
∀n∈N, un+1 =un−vn
2vn=un+vn 2.
(wn)=(un+ivn) (un) (vn)
(un) (u2n) (un)
(un) (u2n) (u3n) (u2n+1) (un)
(un)π(un)
φ1, φ2φ1φ2
(un)
(un)l∈R(uφ1(n)) (uφ2(n))l
(un) (uφ1(n)) (uφ2(n))
(un)n, p ∈N∗0⩽un+p⩽n+p
np
(un) 0
(un)∀n, m ∈N, un+m⩽un+um
un
ninf{un n, n ⩾1} −∞
ε

u0, un+1 = 2un+ 1 u0= 1, un+1 =un+1
2
(un),(vn)
u0= 1, v0= 2 ∀n∈N, un+1 = 3un+ 2vnvn+1 = 2un+ 3vn.
(un−vn)
(un) (vn)
(un) (vn)
2
u0= 1, u1= 0, un+2 −4un+1 + 4un= 0
u0= 1, u1=−1, un+2 −3un+1 +un= 0
u0= 1, u1= 2, un+2 −un+1 +un= 0
u0= 1, u1= 1, un+2 −2cos(θ)un+1 +un= 0
f:R∗
+→R∗ +
∀x > 0, f(f(x)) = 6x−f(x).
a∈R∗
+(un)u0=a un+1 =f(un)
(un) 2
(un)n
f(a)f