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TSI1
no2
n>1an=
n
k=1
1
k
n>1
anxnR
n∈N∗
16|an|=an6n
R1
n>1
znR2
n>1
zn
n
R1>R>R2
R1= 1 R2
αn=xn
n|αn+1|
|αn|=n n+ 1|x|=1
1 + 1
n
|x| −→
n→+∞|x|
|x|<1|x|>1R2= 1 = R1
R= 1
x∈[−1,1] F(x) = 
n>1
anxnx∈]−1,1[
(1 −x)F(x) = (1 −x)
+∞
n=1
anxn=
+∞
n=1
anxn−
+∞
n=1
anxn+1
(1 −x)F(x) =
+∞
n=1
anxn−
+∞
n=2
an−1xn=a1x+
+∞
n=2
(an−an−1)xn
a1= 1 n>2an−an−1=1
n
(1 −x)F(x) =
+∞
n=1
1
nxn
x∈]−1,1[
ln(1 + x) =
+∞
n=1
(−1)n−1
nxn
x(−x) ] −1,1[
ln(1 −x) =
+∞
n=1
(−1)n−1
n(−x)n=
+∞
n=1
(−1)2n−1
nxn=−
+∞
n=1
1
nxn=−(1 −x)F(x)

∀x∈]−1,1[, F (x) = −ln(1 −x)
1−x
+∞
n=0
n!
1×3× ··· × (2n+ 1)x2n+1
n!
1×3× ··· × (2n+ 1) =n!×2×4× ··· × 2n 1×2×3×4× ··· × 2n×(2n+ 1) =(n!)22n (2n+ 1)!
αn=n!
1×3× ··· × (2n+ 1)x2n+1
|αn+1|
|αn|=(n+ 1)! 1×3× ··· × (2n+ 1) ×(2n+ 3)|x|2n+3 ×1×3× ··· × (2n+ 1) n!|x|2n+1
=n+ 1
2n+ 3|x|2−→
n→+∞|x|2
2
lim
n→+∞|αn+1|
|αn|=|x|2 2
αn|x|2
2<1
|x|<√2|x|>√2
R=√2
(E) : (x2−2)y′+xy + 2 = 0
]−√2,√2[
y′+x
x2−2y=2 2−x2
(E) ] −√2,√2[
(H) : y′+x
x2−2y= 0 (x2−2̸= 0)
yH(x) = λe−∫x x
x2−2=λe−1
2ln |x2−2|
(H) ] −√2,√2[ yH:x7→ λ
√2−x2

TSI1
2
√2−x2=2
√21−x2/2=√2
1−(x/√2)2
]−√2,√2[ x7→ 2
√2−x2
x7→ 2 arcsin x
√2+K(K∈R)
(E)ypart(x) = λ(x)
√2−x2
y′
part(x) + x
x2−2ypart(x) = 2 2−x2⇔λ′(x)
√2−x2=2
2−x2
λ′(x) = 2
√2−x2
λ(x) = 2 arcsin x
√2
ypart(x) = 2 arcsin x
√2
√2−x2
(E) ] −√2,√2[ y:x7→ 2 arcsin x
√2+λ
√2−x2
S(x) =
+∞
n=0
(n!)22n
(2n+ 1)!x2n+1 S′(x) =
+∞
n=0
(n!)22n
(2n)! x2n
(x2−2)S′(x) + xS(x) =
+∞
n=0
(n!)22n
(2n)! x2n+2 −
+∞
n=0
(n!)22n+1
(2n)! x2n+
+∞
n=0
(n!)22n
(2n+ 1)!x2n+2
=
+∞
n=1
(n−1)!)22n−1
(2n−2)! x2n−
+∞
n=0
(n!)22n+1
(2n)! x2n+
+∞
n=1
((n−1)!)22n−1
(2n−1)! x2n
=−2 +
+∞
n=1
[2n(2n−1) −4n2+ 2n]((n−1)!)22n−1
(2n)! =−2
S(E) ] −√2,√2[
S(x) = 2 arcsin x
√2+λ
√2−x2λ∈R
S(0) = 0
λ= 0
∀x∈]−√2,√2[, S(x) = 2 arcsin x
√2
√2−x2

I=+∞
0
x
√x(1 + x);J=+∞
0
ln(x)
√x(1 + x)x;K=+∞
0
√xln(x)
(1 + x)2x
I
x7→ 1
√x(1 + x)]0,+∞[
⋆0
1
√x(1 + x)∼
x→0
1
√x
1
0
x
√x1
0
x
√x(1 + x)
⋆+∞1
√x(1 + x)∼
x→+∞
1
x√x∼
x→+∞
1
x3/2
+∞
1
x
x3/2+∞
1
x
√x(1 + x)
I=+∞
0
x
√x(1 + x)
u=√x u = 1/(2√x)x
x7→ √xC1R∗
+R∗ +
I=+∞
0
2u
1 + u2= 2arctan(u)+∞
0=π
I=π
J
x3/4×ln(x)
√x(1 + x)=x1/4ln(x)
1 + x−→
x→00
x1/4ln(x)−→
x→00
ln(x)
√x(1 + x)=
x→0o1
x3/4
c > 0x∈]0, c]
ln(x)
√x(1 + x)
61
x3/4c
0
x
x3/4
c
0
ln(x)
√x(1 + x)
J1=1
0
ln(x)
√x(1 + x)x

TSI1
x5/4×ln(x)
√x(1 + x)∼
x→+∞
ln(x)
x1/4−→
x→+∞0
ln(x)
√x(1 + x)=
x→+∞o1
x5/4
c > 0x>c06ln(x)
√x(1 + x)61
x5/4+∞
c
x
x5/4
+∞
c
ln(x)
√x(1 + x)
J2=+∞
1
ln(x)
√x(1 + x)x
J
J
u= 1/x x = 1/u x =−1/u2u
u7→ 1
uC1]0,1[
]1,+∞[
J1=1
+∞
ln(1/u)
1
√u(1 + 1
u)×−1
u2u=+∞
1
ln(1/u)
1
√u(1 + 1
u)×1
u2u
=+∞
1
−ln(u)
u
√u(u+u
u)u=−+∞
1
ln(u)
√u(1 + u)
J2
J1=−J2
J=J1+J2=−J2+J2= 0
J= 0
K
lim
x→0√xln(x) = 0
lim
x→0
√xln(x)
1 + x= 0
√xln(x)
1 + x∼
x→+∞
ln(x)
√xlim
x→+∞
ln(x)
√x= 0
lim
x→+∞
√xln(x)
1 + x= 0
K
u(x) = √xln(x)v′(x) = 1
(1 + x)2
u′(x) = ln(x)
2√x+1
√xv(x) = −1
1 + x