Séries numériques : Exercices corrigés

un=1
5n
q=1
5
∀n∈N, Sn=
n
X
k=0 1
5k
=1−1
5n+1
1−1
5
=1−1
5n+1
4
5
=5
4× 1−1 5n+1!
−1<1
5<1 lim
n→+∞Sn=5
4
S=5
4
un=−1
3n
q=−1
3
∀n∈N, Sn=
n
X
k=0 −1
3k
=1−−1
3n+1
1 + 1
3
=1−−1
3n+1
4
3
=3
4× 1−−1 3n+1!
−1<−1
3<1 lim
n→+∞Sn=3
4
S=3
4
un=2n
3n−2= 9 ×2 3n

q=2
3
∀n∈N, Sn=
n
X
k=0
9×2
3k
= 9 ×1−2
3n+1
1−2
3
= 9 ×1−2
3n+1
1
3
= 27 × 1−2
3n+1!
−1<2
3<1 lim
n→+∞Sn= 27
S= 27
un=tan π
7n
3n+2 =1 9× tan π
7
3!n
q=tan π
7
3
∀n∈N, Sn=
n
X
k=0
1
9× tan π
7
3!k
=1
9×
1−tan(π
7)
3n+1
1−tan(π
7)
3
=1
91−tan(π
7)
3×
1− tan π
7
3!n+1
−1<tan π
7
3<1 lim
n→+∞Sn=1
91−tan(π
7)
3
S=1
91−tan(π
7)
3=1
9−3 tan π
7
un=9
(3n+ 1)(3n+ 4)
n un∼
n→+∞
9
9n2=1 n2
1
n2α= 2 >1
Pun

∀n∈N∗,
n
X
k=0
9
(3k+ 1)(3k+ 4) =
n
X
k=0 3
3k+ 1 −3 3k+ 4
=
n
X
k=0
3
3k+ 1 −
n
X
k=0
3
3k+ 4
=
n
X
k=0
3
3k+ 1 −
n
X
k=0
3
3(k+ 1) + 1
=
n
X
k=0
3
3k+ 1 −
n+1
X
k=1
3
3k+ 1
=3
3×0+1+
n
X
k=1
3
3k+ 1 − n
X
k=1
3
3k+ 1 +3 3n+ 4!
= 3 +
n
X
k=1
3
3k+ 1 −
n
X
k=1
3
3k+ 1 −3 3n+ 4
= 3 −3
3n+ 4
lim
n→+∞3−3
3n+ 4= 3
+∞
X
k=0
un= 3
X1
2n=1 2X1 n
1
nX1 2n
un=1
2n+ 1
n un∼
n→+∞
1
2n
1
2n
X1
2n+ 1
un= ln 1−1
n2
n un∼
n→+∞−1
n2lim
n→+∞
1
n2= 0
−1
n2α= 2 >1
Pun

∀n∈N,
n
X
k=2
ln 1−1
k2=
n
X
k=2
ln k2−1
k2
=
n
X
k=2
ln (k+ 1)(k−1)
k2
=
n
X
k=2
(ln(k+ 1) + ln(k−1) −2 ln(k))
=
n
X
k=2
ln(k+ 1) +
n
X
k=2
ln(k−1) −2
n
X
k=2
ln(k)
=
n+1
X
k=3
ln(k) +
n−1
X
k=1
ln(k)−2
n
X
k=2
ln(k)
=
n−1
X
k=3
ln(k) + ln(n) + ln(n+ 1)
+ ln(1) + ln(2) +
n−1
X
k=3
ln(k)
−2 ln(2) −2
n−1
X
k=3
ln(k)−2 ln(n)
= ln(n+ 1) −ln(n)−ln(2)
= ln n+ 1
n−ln(2)
lim
n→+∞ln n+ 1
n−ln(2)=−ln(2)
+∞
X
n=2
un=−ln(2)
un=n2+ 1
n2lim
n→+∞un= 1
0Pun

un=2
√nα=1
2<1
un=(2n+ 1)4
(7n2+ 1)3
n un∼
n→+∞
16n4
343n6=16 343n2
16
343n2α= 2 >1
Pun
un=1−1
nn
= enln(1−1
n)
ln 1−1
nnlim
n→+∞
1
n= 0
∀n∈N, un= e
n−1
n+o
n→+∞(1
n)= e−1+ o
n→+∞
(1) lim
n→+∞un= e−16= 0
0Pun
un=ne1
n−n=ne1 n−1
e1
nnlim
n→+∞
1
n= 0
∀n∈N, un=n1 + 1
n+o
n→+∞1
n−1=n1 n+o
n→+∞1
n= 1 + o
n→+∞(1)
lim
n→+∞un= 1 6= 0
0Pun
un= ln(1 + e−n)
nlim
n→+∞e−n= 0 un∼ n→+∞e−n=1
en
1
en
−1<1
e<1
Pun
un=n
n+ 1n2
= en2ln(n
n+1 )= e−n2ln(n+1 n)= e−n2ln(1+ 1 n)
ln 1 + 1
nn
lim
n→+∞
1
n= 0 ∀n∈N
un= e−n21
n−1
2n2+o
n→+∞(1
n2)= e−n+1
2+o
n→+∞
(1) =1
en
e
1
2+o
n→+∞
(1)
2n2
un∼
n→+∞1
en
e1
21
en
e1
2
−1<1
e<1Pun