[pdf]

Pn≥0n2+1
3nzn
Pn≥0e−n2znPn≥1ln n
n2z2n
Pn≥0nn
n!z3n
Pn≥0n!zn
Pn≥02n
nznPn≥0
(3n)!
(n!)3zn
Pn≥0n+1
√n+ 1 −n
√nzn
Pn≥0zn2Pn≥0sin(n)znPn≥1
sin(n)
n2zn
Xln n+ 1
nxnXsin(e−n)xn
Panxn(an)
a0=α, a1=β∀n∈N, an+2 = 2an+1 −an
(α, β)∈R2
Pπ√n2+2nx2n
ann√3
P+∞
n=1 anxn
α∈RPn≥1
cos(nα)
nxn
X
n≥1
d(n)znX
n≥1
s(n)zn
d(n)s(n)
n
α Rα
X
n≥1
xn
sin(nπα)
Rα≤1
(un)n≥1
u1= 2 ∀n≥1, un+1 = (un)un
n≥1
un
un+1 ≤1 (n+ 1)n
1/un
α=
+∞
X
n=1
1
un
α

C
n≥1
πun
+∞
X
k=n+1
1
uk≤C
uun−1
n
Rα= 0
α
Pn≥0anznR z0∈C
Pn≥0anzn
0R
n
p|an| → `∈R+∪ {+∞}
Panzn
α∈RPanznPnαanzn
PanznR
Panz2n
PanznR
Xa2
nzn
PanznR > 0
Xan
n!zn
Panzn
r > 0|an| ≤ 1/rn
Pan
n!zn
Sn=Pn
k=0 ak
PSn
n!zn
(an)
XanznX1
an
zn
PanznR
bn=an
1 + |an|
R0Pbnzn
R0≥max(1, R)
R0>1R0=R
R0R
PanznPbnznRaRb
n∈Nanbn= 0
P(an+bn)znR= min(Ra, Rb)

n∈N∗
In=Z+∞
1
e−tndt
(In)
(In)
R
InxnR−R
p, q ∈N
I(p, q) = Z1
0
tp(1 −t)qdt
I(p, q)
un=I(n, n)
Punxn
(fn)
∀n≥2,∀x∈R, fn(x)=(−1)nln(n)xn
Pfn
S
∀x∈]−1 ; 1[, S(x) = 1
1 + x +∞
X
n=1
(−1)n+1 ln 1 + 1
nxn+1!
S1−
lim
x→1−
S(x) = 1
2 +∞
X
n=1
(−1)n+1 ln 1 + 1
n!
lim
n→+∞
1×3× ··· × (2n−1)
2×4× ··· × (2n)√n=1
√π
(an)
an+1 = ln(1 + an)a0>0
Panxn
(Panxn)
1/an+1 −1/an
x
f(x) =
+∞
X
n=1
xn
√n
R f
−1
f−1
f
I s x
s(x) =
+∞
X
n=1
xn
√n
s0I∩R+
s I ∩R+
(1 −x)s0(x)s0I
fR+
f(x) = √x+ 1 −√xx
s

f:x7→
+∞
X
n=1
sin 1
√nxn
R f
−R R
f(x)x→1−
x→1−
(1 −x)f(x)→0
∀z∈C, c(z) =
+∞
X
n=0
(−1)n
(2n)! z2ns(z) =
+∞
X
n=0
(−1)n
(2n+ 1)!z2n+1
∀z∈C, c(z)2+s(z)2= 1
PanznR > 0f
P+∞
n=0 a2nz2nf|z|< R
P+∞
n=0 a3nz3n
(an)T T ∈N∗
Pn≥0anxn
PnT −1
k=0 akxkP+∞ n=0 anxn
x∈]−1 ; 1[ x
(an)Sn=Pn
k=0 ak
Sn→+∞an/Sn→0
Pn≥0anxnPn≥0Snxn
S(x) = P+∞
n=0 anxnR > 0
α > 0 [0 ; α]S(x)=0
S= 0
P∞
n=0 anznR > 0
f(z)
0< r < R
∞
X
n=0 |an|2r2n=1
2πZ2π
0f(reiθ)
2dθ
f|f|R= +∞P∈RN[X]
|f(z)| ≤ P(|z|)z f ∈CN[X]
B={z∈C,|z| ≤ 1}f B C
B◦
(Pk)k≥0f B
I x
+∞
X
n=1
ln(n)xn
f(x)

I
a1=−1an=−ln 1−1
n−1 nn≥2
g:x7→
+∞
X
n=1
anxn
f g
f(x)x→1−
f(x)x→ −1+
x→1−
f(x) =
+∞
X
n=0
xn2
(an)Panxn
R
X(anln n)xnX an
n
X
k=1
1
k!xn
P+∞
n=1 ln(n)xnx→1−
+∞
X
n=1
ln1 + 1
nxn
f(x) =
+∞
X
n=1
ln 1 + 1
nxn
f(−1) R1
0
(−1)E(1/x)
xdx E
f x = 1
an=Z+∞
n
th t
t2dt
n∈N∗
P+∞
n=1 anxnx
f(x)
f−1
f1−
PanxnR= 1
x∈]−1 ; 1[
S(x) =
+∞
X
n=0
anxn
(an)S
[0 ; 1[
Pan
lim
x→1− +∞
X
n=0
anxn!=
+∞
X
n=0
an
PanxnR= 1
∀n∈N, an≥0