Exercice 1 Exercice 2 Exercice 3

∀(n, k)∈(N∗)2k≤n
k(n
k) = nn−1
k−1
∀n∈NSn
Sn=
n
X
k=0
k(n
k)
∀k∈N∗
1
k+ 1 ≤ln(k+ 1) −ln(k)≤1 k
(un)n∈N∗∀n∈N∗
un=
n
X
k=1
1
k
∀n∈N∗un≥ln(n+ 1)
(vn)n∈N∗∀n∈N∗
vn=un−ln(n)
∀n∈N∗vn≥0
(vn)n∈N∗
vnγ
(wn)n∈N∗∀n∈N∗
wn=
n−1
X
k=0
1
n+k
(wn)n∈N∗
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∈N∗
Pn=Y
1≤i<j≤n
ij
(un)n∈N∗α
(vn)n∈N∗∀n∈N∗
vn=Pn
i=1 ui
n
α
lim un= +∞lim vn= +∞
lim un=−∞ lim vn=−∞
(un)n∈N∗∀n∈N∗
un=1
n+1 2n+. . . +1 n×n
α∈R\ {0,4}(un)n∈N∗
u0∀n∈N, un+1 = 4un−2n2+ 3n+1+αn
a, b, c (wn)n∈N∀n∈N
wn=an2+bn +c
∀n∈N
wn+1 = 4wn−2n2+ 3n+ 1
β(tn)n∈N∀n∈N
tn=βαn
∀n∈N
tn+1 = 4tn+αn
unu0α n

∀n∈N∗
n
Y
k=1
(2k+ 1) = (2n+ 1)!
2nn!
(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
f:R+−→ R
∀n∈N
un=f(n)vn=
n
X
k=0
uk−Zn
0
f(x)dx
(vn)n∈N