Séries entières 1. Déterminer le rayon de

2015 −2016
X3
√nznXzn
3n5+ 1 X7n n4+ 3znXth(n)znXcos n nαznX1 ln(n!)zn
x
cos(2x) = 2 cos2(x)−1
x(x−1)y00 −xy? + y= 0
Xn
(2n+ 1)!xn
Xch(n)
nxn
ΣanznR > 0
Σanz3nX|an|
1 + |an|znX|an|4
5zn
α∈RPArctan(nα)xn
PanznR > 0
Xan
n!zn
a0>0∀n∈N
an+1 = ln(1 + an)Panzn
Xxn
(n2+ 7n+ 12) X3n n+ 2xnX2n+ 1 2n−1xnXn (2n+ 1)!xn
n an=cos 2nπ
3
n+ cos 2nπ
3
Panxn
X1 + 1
2+··· +1 nxn
Xln(n)xn

2015 −2016
+∞
X
n=0
n−3
n!xn
+∞
X
n=0
(n+ 1)(n−2)
n!xn
+∞
X
n=1
x2n
n(2n+ 1)
+∞
X
n=0
(−1)n+1 x2n+1
4n2−1
a, b 0< a < b (un)
n un=anun=bn
Punxn
x7→ ln(x2−5x+ 6) x7→ arctan xsin α
1−xcos α, α 6≡ 0[π]x7→ e−xsin x
0
f f(x) = Arctan √2x
1−x2!
X3nxn
n+ 2
Σanzn
an=1 + 1
nn2
an=
n
Y
k=0 1 + 1
k+ 1an=Z2n
n
et
t2+ 1dt
ann3
X
n>1
(−1)ne−nx
n
F(x) = Zx
0
e−tcos(t)dt
Z1
0
e−tcos tdt
n In=Z1
0
un(1 −u)ndu
X(n!)2
(2n+ 1)!xn
p Sp(x) =
+∞
X
n=0 n+p
pxn
(1 −x)Sp0(x)
(an)n∈N
PanznPa2
nzn
Xxn
4n2−1
x7→ sin (αArc sin (x)) α∈R
f c∞]−R, R[R > 0
∀n∈Nf(n)[0, R]
Xf(n)(0)
n!xn]−R, R[