Exercices corrigés sur les séries entières

anzn
an= ln n, an= (ln n)n, an= (√n)n, an=en1/3, an=nn
n!, an= arcsin n+ 1
1 + n√2−π
4.
an
1 + bnzn
a, b ∈R∗
+
anzn
an=n n
0an=2n∃k∈N:n=k3 0
Rn≥0anxn
]−R, R[
an=n, an=n(n−1), an=n2.
Rn≥1anxn
]−R, R[
an=1
n(n+ 2), an=1 ncos 2nπ 3.
∞
n=0
n
2n,
∞
n=0
n2
2n,
∞
n=0
n2−3n+ 2
2n.
f(x) = 1
(x−1)(x−2), g(x) = ln(x2−5x+ 6), h(x) = x
0
cos t2dt.
t
f(x) = 1
x2−2tx + 1.

f(x) = arctan xsin a
1−xcos aa∈]0, π[, g(x) = arcsin x2.
fp(x) := 1
(x−1)(x−2) ···(x−p).
fp(x) =
p
k=1
λk
x−kλk= (−1)p−kk p!Ck
p.
fp
a0= 1 b0= 0 n∈N
an+1 =−an−2bnbn+1 = 3an+ 4bn.
n≥0
an
n!xn
n≥0
bn
n!xn.
3xy′+(2−5x)y=x
x∈]−π/2, π/2[ f(x) = tg x
f′= 1 + f2(Pn)
Nf(n)=Pn◦f n ∈N
f
R π/2
angan
n≥1
(n+ 1)an+1 =
n
k=0
akan−k.
x∈]−π/2, π/2[ f(x) = g(x)R=π/2
a0, a1, . . . , a7
x7→ th x
an

•
an+1
an
=ln(n+ 1)
ln n=ln n(1 + n−1) ln n= 1 + ln(1 + n−1) ln n−→ 1n→ ∞,
R= 1
• |an|1/n = ln n→ ∞ n→ ∞ R= 0
• |an|1/n =√n→ ∞ n→ ∞ R= 0
• |an|1/n =en1/3n−1=en−2/3→1n→ ∞
R= 1
•an+1
an
=(n+ 1)n+1
(n+ 1)!
n!
nn=n+ 1 nn
=1 + 1
nn
−→ e n → ∞,
ln 1 + 1
nn
=nln 1 + 1
n∼n1 n= 1.
R= 1/e
•n+ 1
1 + n√2=1 + n−1
√2 + n−1
=1
√2
1 + n−1
1 + (n√2)−1
∼1
√21 + 1
n1−1
n√2
∼1
√21 + 1
n1−1
√2
∼1
√2+1 √2−1
21 n.
arcsin 1/√2 arcsin′(x) = (1−x2)−1/2
arcsin′(1/√2) = √2
an= arcsin n+ 1
1 + n√2−π
4
∼arcsin 1
√2+1 √2−1
21 n−arcsin 1
√2
∼arcsin′1
√2·1 √2−1
21 n
=√21
√2−1
21 n
=1−1
√21
n=: αn.
(αn)1/n →1n→ ∞ R= 1

a= 0 an≡0R=∞a̸= 0
•b > 1
an
1 + bn∼n→∞ a bn,an 1 + bn1/n
−→ a
bn→ ∞.
R=b/a
•b= 1
an
1 + bn=an 2,an 1 + bn1/n
=a
21/n −→ a n → ∞.
R= 1/a
•b < 1
an
1 + bn∼n→∞ an,an 1 + bn1/n
−→ a n → ∞.
R= 1/a
•|an|1/n =n1/n n
0
ln n1/n =1
nln n−→ 0, n1/n −→ e0= 1 n→ ∞.
lim sup |an|1/n = 1 R= 1
•|an|1/n =2∃k∈N:n=k3
0
lim sup |an|1/n = 2 R= 1/2
•an+1
an
=n+ 1
n−→ 1, R = 1.
∞
n=0
nxn=x
∞
n=1
nxn−1=x
(1 −x)2,
n≥1nxn−1
∞
n=0
xn=1
1−x, x ∈]−1,1[ .

•an+1
an
=(n+ 1)n
n(n−1) −→ 1, R = 1.
∞
n=0
n(n−1)xn=x2
∞
n=2
n(n−1)xn−2=2x2
(1 −x)3,
n(n−1)xn−2
∞
n=0
xn=1
1−x, x ∈]−1,1[ .
•an+1
an
=(n+ 1)2
n2−→ 1, R = 1,
∞
n=0
n2xn=
∞
n=0 n(n−1) + nxn
=
∞
n=0
n(n−1)xn+
∞
n=0
nxn
=2x2
(1 −x)3+x (1 −x)2
=x2+x
(1 −x)3.
•an+1
an
=n(n+ 2)
(n+ 1)(n+ 3) −→ 1n→ ∞,
R= 1
1
n(n+ 2) =1 21 n−1 n+ 2,
∞
n=1
xn
n(n+ 2) =1 2∞
n=1
xn
n−
∞
n=1
xn
n+ 2.
∞
n=1
xn
n=
∞
n=0
xn+1
n+ 1 =x
0∞
n=0
tndt=x
0
1
1−tdt=−ln(1 −x),
x̸= 0
∞
n=1
xn
n+ 2 =x−2
∞
n=1
xn+2
n+ 2 =1 x2∞
n=1
xn
n−x−x2 2=−1 x2ln(1 −x) + x+x2 2.
x= 0 x̸= 0
∞
n=1
xn
n(n+ 2) =1 21 x2−1 2ln(1 −x) + 1 x+1 2.