Stolz Cesaro Lý thuyết và một số ứng dụng

The purpose of this note is to show how
obtain limit or equivalent of sequence with Theorem Stolz-Cesaro
by Moubinool OMARJEE

Let me remind the main tool in this note
___________________________________________________________
Theorem of Stolz-Cesaro: a n  n≥0 , b n  n≥0 sequences of reals numbers.
Assume b n  n≥0 is strictly increasing , unbounded and the following
limit exist
lim a n+1 − a n = L
n→+∞ b n+1 − b n
Then the
lim a n also exist and its equal to L
bn

n→+∞

____________________________________________________________
APPLICATION 1 : x n  n≥1 sequence of real number with x n > 0 for any n ≥ 1
lim x n = +∞
n→+∞ n
Then
n

∑

lim 1
n→+∞
n

k=1

1 =0
xn

Proof: With S.C (Stolz-Cesaro)
n

lim 1
n→+∞
n

∑
k=1

1 = lim
n→+∞
xn

1
x n+1

= n→+∞
lim

n+1 − n

n+1 + n
n+1
= n→+∞
lim 2

=0
x n+1
x n+1

APPLICATION 2 : x n  n≥1 sequence of real number given by
xn =

n
k. k!
∑ k=1

n + 1! − 1

Find the limit of x n  n≥1
Proof: With S.C
lim x
n→+∞ n

= n→+∞
lim

n + 1n + 1!
n + 1n + 1!
= n→+∞
lim
=1
n + 2! − n + 1!
n + 1n + 1!

APPLICATION 3 :x n  n≥1 sequence of real number with x 1 = 1

x n+1 = x 1 + x 2 +. . . +x n
Find
lim x n
n→+∞ n
Proof: All the term of x n  n≥1 are strictly positive x 2n+1 = x 2n + x n
x n  n≥1 is strictly increasing, if it was bounded then
lim x
n→+∞ n

exist = L

L2 = L2 + L
This gives the limit L = 0 contradiction since all x n ≥ 1. So


lim x
n→+∞ n

= +∞

x 2n+1
= 1 + x1n → 1
x 2n
lim x n+1 = 1
n→+∞ x n
With S.C theorem
x n = lim x n+1 − x n = lim x 2n+1 − x 2n
lim
n→+∞ n
n→+∞ n + 1 − n

n→+∞ x n+1 + x n
= n→+∞
lim x x n+ x = n→+∞
lim x n+11
= 1 = 1
n+1
n
1+1
2
+
1
xn
APPLICATION 4 :Find
lim

n→+∞

lnn! − n lnn
n

Proof: With S.C.
lnn! − n lnn
lnn + 1 − n + 1 lnn + 1 + n lnn
lim
= n→+∞
lim
n
n→+∞
n + 1 − n
= n→+∞

lim −n lnn + 1 + n lnn = n→+∞
lim −

ln1 +

1
n

1
n



= −1

APPLICATION 5: Find
n
Ene n 
∑ k=1
1

lim

n2

n→+∞

Proof: With S.C
n
Eke k 

∑ k=1
1

lim

n→+∞

n2

1

Ene n 
n
= n→+∞
lim 2
= n→+∞
lim
= 1
2
2n + 1
n − n − 1 2

APPLICATION 6: x n  n≥1 sequence of real number with
n

lim x
x2 = 1
n→+∞ n ∑ k
k=1

Prove that
lim 3 3n x n = 1

n→+∞

Proof: S n = ∑ k=1 x 2k , x n S n  tend to 1. The sequence S n  is strictly increasing
n
if S n  converge to L , then x n → L1 > 0 and we will have S n = ∑ k=1 x 2k → +∞
contradiction. We have
n

lim S n = +∞ and n→+∞
lim x n = 0

n→+∞

We also have x n S n−1 → 1 when n → +∞ and
S 3n − S 3n−1 = x 2n S 2n + S n S n−1 + S 2n−1  → 3 when n → +∞
With S.C.
S 3 − S 3n−1
S3
lim n
=3
lim nn = n→+∞
n − n − 1

n→+∞

We deduce that

3

3

3n
→1
Sn

3n x n =

3

3n
x S →1
Sn n n

APPLICATION 7: x n  n≥1 sequence of reals numbers with 0 < x 1 < 1 and
x n+1 = x n − x 2n for any n ≥ 1. Prove that
lim nx n = 1

n→+∞

Proof: The sequence x n  n≥1 is decreasing and bounded
lim x
n→+∞ n

= L exist

L = L − L2

So L = 0
Consider y n =

1
xn

with S.C.
y
y − yn
x n+1
lim n = n→+∞
lim n+1
= n→+∞
lim x nx n−x n+1
n→+∞ n
n + 1 − n
x 2n
1
= n→+∞
lim
= n→+∞
lim
=1
1 − xn
x n x n − x 2n 

APPLICATION 8: x n  n≥1 sequence of reals numbers with x 1 = 1
x n+1 =

x 2n +2

n+1

for any n ≥ 1. Find
n
xk
∑ k=1
lim
1
n→+∞ ∑ n
k=1 k

Proof: By induction 0 < x n ≤ 2 for any n ≥ 2
2
0 < x n+1 ≤ 2 + 2 → 0 when n → +∞
n+1

With S.C.
n
xk
∑ k=1
x n = lim n x 2n−1 + 2 = 2
lim
=
lim
1
n
n→+∞ ∑ n
n→+∞ 1
n→+∞
n

k=1 k

APPLICATION 9: x n  n≥1 sequence of reals numbers such that
lim x n+1n− x n = L > 0
n→+∞
Prove that
lim

n→+∞

n+1

x n+1
− xn
n n!
n + 1!

= eL
2

Proof: Since x n+1 − x n ∼ nL the equivalence of partial sum for divergence
n−1
n−1
gives ∑ k=1 x k+1 − x k  ∼ L ∑ k=1 k this gives xnn ∼ nL
2
With S.C.
x n+1
− nx n
n+1 n+1!
n!

x n+1
lim
− xn
= n→+∞
lim
n→+∞
n n!
n + 1 − n
n+1 n + 1!
= n→+∞
lim

x n = eL
2
n n n!


Since

n

1
n!

=

e
n

−

e lnn+ln2π
2
n2

+O

1
n3