Exercice 1 Exercice 2 Exercice 3 Exercice 4

(un)n∈N∀n∈N0≤un≤1un−→ 1
∀n∈N∗
n
X
k=1
k3= n
X
k=1
k!2
(un)n∈N
u0=2
3∀n∈N, un+1 =un 2+n
2√2+1
√2
(vn)n∈N
∀n∈N, vn=un√2−n
(vn)n∈N
n vnun(un)n∈N
(sn)n∈N
∀n∈N, sn=
n
X
i=0
ui
snn
n≥1x∈R∗
+
Pn(x) = −1 +
n
X
k=1
xk
Pn(x)=0 xnR∗
+
0< xn≤1
(xn)n∈N∗`
lim
n−→+∞(xn)n+1 = 0
Pn(x)
`=1
2
∀(n, k)∈(N∗)2k≤n
k(n
k) = nn−1
k−1
∀n∈NSn
Sn=
n
X
k=0
k(n
k)

∀n∈Nun+1 > unun−→ +∞
(un)n∈N(eun)n∈N
∀x1, . . . , xn, y1, . . . , yn
n
X
k=1
xkyk= n
X
k=1
xk! n
X
k=1
yk!
v∈R0< v < 1 (un)n∈N
0< u0< u1∀n∈N∗, un+1 =un+vnun−1
∀n≥2
un≤u1
n−1
Y
k=1
(1 + vk)
(un)n∈N
(wn)n∈N(tn)n∈N
∀n∈N, tn=w0+ 2w1+. . . + 2nwn
2n+1
lim wn=αlim tn=α
(an)n∈N
a0= 1 ∀n≥0, an+1 =1+(n+ 1)an
2n+ 3
∀n∈N
0< an≤2
n+ 1
(an)n∈N
(un)n∈N∀n≥0un=nan(vn)n∈N∀n≥0
vn= 2un+1 −un
(vn)n∈N
∀n≥0
un+1 =v0+ 2v1+. . . + 2nvn
2n+1
(un)n∈N
∀(n, k, i)∈N3k≤i≤n
(n
i)i k= (n k)n−k
n−i
∀n∈NSn
Sn=
n
X
k=0
n
X
i=k
(n
i)i k

limn(un+1 −un) = 0 (un)n∈N
(un)n∈Nun+1 = (n+ 1)unn∈Nu0= 1
∀n∈Nun= (n+ 1)n
X
1≤i<j≤n
xi,j
n(n−1)
2
x∈]0,1[
ln(1 + x)≤x≤ −ln(1 −x)
(un)n∈N∗
∀n≥1, un=
2n
X
k=n
1
k
(φn)n∈N
φ0= 0, φ1= 1 ∀n≥2, φn+2 =φn+1 +φn
φnn
∀n∈N
φ2
n+1 −φnφn+2 = (−1)n
φn+1
φnn∈N
n∈N
n
X
k=0
(n
k)φk=φ2n
n∈N
n
X
k=0
(n
k) (−1)kφk=−φn
n≥2
In=
n
Y
i=2 1−1
i2
n≥1
Jn=
n
Y
i=1
√i
2i