Séries Numériques : Exercices et Problèmes Corrigés

un=n
n2+1
un=ch(n)
ch(2n)
un=1
√n2−1−1 √n2+1
un= e −1 + 1
nn
un=n
n+1 n2un=1
ncos2nun=1
(ln n)ln n
un=1
n1+ 1
n
e−1 + 1
nn
n3/2− bn3/2c+n
(un)n≥1(vn)
vn=1
n(n+ 1)
n
X
k=1
kuk
PunPvn
+∞
X
n=1
un=
+∞
X
n=1
vn
Pan
(bn)
Xan
bnXbn
α∈R
un= e−nαun=ln n
nαun= exp(−(ln n)α)
a, b ∈R
X
n≥1ln(n) + aln(n+ 1) + bln(n+ 2)
a, b, c
a
√1+b √2+c √3+a √4+b √5+c √6+. . .
λ
un=λn
1 + λ2n, vn=λ2n 1 + λ2n, wn=1 1 + λ2n

α∈Rf∈ C0([0 ; 1],R)f(0) 6= 0
un=1
nαZ1/n
0
f(tn) dt
α > 0 (un)
u1/n
n= 1 −1
nα+ o1 nα
un
a > 0
un=a(a+ 1) . . . (a+n−1)
n!
un
x∈R∗
+n∈N∗
un=n!
xn
n
Y
k=1
ln1 + x
k
ln(un+1)−ln(un)
(un)n≥1
α∈R
ln(un+1)−ln(un)−αln1 + 1
n
A∈R∗un∼Anα
un
(un)n∈N(vn)n∈N
vn=u2n+u2n+1
Xun⇐⇒ Xvn
PunPvn
Xmax(un, vn),X√unvnXunvn
un+vn
Pun
X√unun+1
a
an
√anan+1
(un)
n∈N
vn=un
1 + un
PunPvn
vn=un
u1+··· +un
ln(1 −vn)

(un)n≥0(vn)n≥0(R+)N
∀n∈N, vn=1
1 + n2un
vn
un
Pan
Pa1−1/n
n
Pan
un=a0a1. . . an
(un)
n
X
k=1
uk−nun
un
X
n≥1
1
n(n+ 1)(n+ 2)
P+∞
k=1 1
k2=π2 6
+∞
X
k=1
1
k2(k+ 1)2
P+∞
n=0 1
n!= e
+∞
X
n=0
n+ 1
n!
+∞
X
n=0
n2−2
n!
x∈]−1 ; 1[
+∞
X
n=1
xn
(1 −xn)(1 −xn+1)
m≥1
Sm=
+∞
X
n=1
1
n(n+ 1) . . . (n+m)
p∈N
ap=
+∞
X
n=0
np
2n
apapa0, . . . , ap−1
ap∈N

α∈]2 ; +∞[ (an)
a0=α an+1 =a2
n−2n∈N
+∞
X
n=0
1
a0a1. . . an
=1
2α−pα2−4
(un)n∈N(vn)n∈N
un+1
un≤vn+1 vn
un=
n→+∞O(vn)
un+1
un
=
n→+∞1−α
n+ o1 nα > 1
Pun
un+1
un
=
n→+∞1−α
n+ o1 nα < 1
Pun
(un)n≥0(vn)n≥0λ∈R
∀n∈N, un≥0,X|vn|un+1
un
= 1 −λ
n+vn
(nλun)
nn
n!en
un=1
3nn!
n
Y
k=1
(3k−2) vn=1
n3/4
n
un+1
un≥vn+1 vn
Punun
vn
(un)
vn=un+1
Sn
Sn=
n
X
k=0
uk
PunPvn
(un)n≥1
n∈N∗
vn=un/SnSn=u1+··· +un
Pvn
(un)
un
n
Rn=
+∞
X
k=n+1
uk
un/Rnun/Rn−1

(an)Sn=Pn
k=0 ak
PanPan/Sn
Pan
∀n∈N∗,an
S2
n≤1
Sn−1−1 Sn
Pan/S2
n
Pan
Pan/Sn
(un)α
Xnαun
nα+1un→0
X
n≥2
1
nln n
a∈]0 ; 1[ Pn≥0a√n
α > 1
ζ(α) =
+∞
X
n=1
1
nα
(α−1)ζ(α)α1+
Pn
k=1 √k∼2
3n√nln(n!) ∼nln nPn
k=2 1
kln k∼
ln(ln n)
lim
a→+∞
+∞
X
n=1
a
n2+a2
a > 0b > 0n∈N∗
An=1
n
n
X
k=1
(a+bk), Bn=
n
Y
k=1
(a+bk)1/n
limn→+∞Bn
Ane
nα
Pn
k=2 ln2k
α