Download C.2 The number e

C.2
The number e
Pag. 72 ←− Properties of the sequence an =
1+
1
n
n
We shall prove some properties enjoyed by Euler’s number.
Property C.2.1 The sequence {an } is strictly increasing.
Proof. Using Newton’s formula (1.13) and (1.11), we may write
an =
1
1+
n
n
n n
X
X
n 1
1 n(n − 1) · · · (n − k + 1)
=
=
k nk
k!
nk
k=0
k=0
n
X
n−k+1
1 nn−1
···
=
k! n n
n
k=0
n
X 1
1
k−1
·1· 1−
=
··· 1 −
;
k!
n
n
(C.2.1)
k=0
similarly,
an+1 =
n+1
X
k=0
1
1
k−1
·1· 1−
··· 1 −
.
k!
n+1
n+1
We note that
1
1
1−
< 1−
,
n
n+1
···
(C.2.2)
k−1
k−1
, 1−
< 1−
,
n
n+1
so each summand of (C.2.1) is smaller than the corresponding term in (C.2.2).
The latter sum, moreover, contains an additional positive summand labelled by
k = n + 1. Therefore an < an+1 for each n.
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C.2 The number e
Property C.2.2 The sequence {an } is bounded; precisely,
2 < an < 3 ,
∀n > 1 .
Proof. Since a1 = 2, and the sequence is strictly monotone by the previous
property, we have an > 2 , ∀n > 1.
Let us show that an < 3 , ∀n > 1. By (C.2.1), and observing that k! = 1 · 2 ·
3 · · · k ≥ 1 · 2 · 2 · · · 2 = 2k−1 , it follows
X
n
n
X
k−1
1
1
1
··· 1 −
<
·1· 1−
k!
n
n
k!
k=0
k=0
n
n
n−1
X
X
X 1
1
1
= 1+
≤1+
=1+
.
k!
2k−1
2k
an =
k=1
k=1
k=0
Example 5.27 will tell us that
n−1
X
k=0
1 − 21n
1
1
< 2.
=
2
1
−
=
2k
2n
1 − 21
We conclude that an < 3.
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Property C.2.3 Euler’s number e is irrational, and lies between 2 and 3.
Proof. Based on the First comparison theorem for sequences (p. 137, Theorem
4.), from the previous property we quickly deduce
2 < e ≤ 3.
(C.2.3)
Suppose, by contradiction, that e is a rational number, so that there exist two
m0
integers m0 and n0 6= 0 such that e =
. Recall that for any n ≥ 0
n0
e=
n
X
ex̄n
1
+
,
k! (n + 1)!
0 < x̄n < 1
k=0
(see Remark 7.4). From this,
n
n!e = n!
X n!
m0
ex̄n
=
+
.
n0
k! n + 1
k=0
As the exponential map is monotone, and using (C.2.3), we deduce
1 = e0 < ex̄n < e < 3 .
(C.2.4)
C.2 The number e
3
n
X
m0
n!
Choosing now n ≥ max(3, n0 ), the numbers n!
and
are integers,
n0
k!
k=0
ex̄n
whereas
lies in the open interval between 0 and 1. The identity (C.2.4) then
n+1
must be false, so e is irrational and equality in (C.2.3) never occurs.
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Pag. 105 ←− Proof of Property 4.20
Property 4.20 The following limit holds
x
1
= e.
lim
1+
x→±∞
x
Proof. We start by considering the limit for x → +∞. Denoting by n = [x] the
integer part of x (see Examples 2.1), by definition n ≤ x < n + 1; from that it
1
1
1
1
1
1
< ≤ , in other words 1 +
< 1 + ≤ 1 + . The familiar
follows
n+1
x
n
n+1
x
n
features of power functions yield
n x x x n+1
1
1
1
1
1
1+
≤ 1+
< 1+
≤ 1+
< 1+
,
n+1
n+1
x
n
n
hence
1+
1
n+1
n+1 1
1+
n+1
−1
<
1
1+
x
x
<
1
1+
n
n 1
1+
n
. (C.2.5)
When x tends to +∞, n does the same. Using (3.3) we have
n n
1
1
1
1
lim
lim
1+
= lim
= e;
1+
1+
1+
n→+∞
n→+∞
n→+∞
n
n
n
n
the substitution m = n + 1 similarly gives
n+1 −1
1
1
lim
1+
= e.
1+
n→+∞
n+1
n+1
Applying the Second comparison theorem 4.5 to the three functions in (C.2.5)
proves the claim for x → +∞. Now let us look at the case when x tends to −∞.
If x < 0 we can write x = −|x|, so
x −|x| −|x| |x| |x|
1
1
1
|x| − 1
|x|
1+
= 1−
=
=
= 1+
.
x
|x|
|x|
|x| − 1
|x| − 1
Set y = |x| − 1 and note y tends to +∞ as x goes to −∞. Therefore,
x
y+1
y
1
1
1
1
lim
1+
= lim
1+
= lim
1+
lim
1+
= e.
x→−∞
y→+∞
y→+∞
y→+∞
x
y
y
y
This concludes the proof.
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