De…ne Euler’s number e by the series
1
X
1
e=
:
n!
n=0
Let an =
1
. Then
n!
an+1
1
= lim sup
= 0:
an
n
+
1
n!1
lim sup
n!1
P1 1
Hence, by the Ratio Test, n=0
converges. At the top of p. 64 Rudin shows
n!
that e < 3. The …rst 50 digits of e are
2:7182818284590452353602874713526624977572470936999:
Theorem 3.31.
e = lim
1+
n!1
1
n
n
:
Proof. We de…ne two sequences:
sn =
n
X
1
k!
and tn =
1+
k=0
1
n
n
:
By de…nition, we have
lim sn =
n!1
1
X
1
= e and
n!
n=0
lim tn = lim
n!1
1+
n!1
1
n
n
:
So the theorem will follow from proving
lim sup tn
e and
lim inf tn
(1) lim supn!1 tn
n
(a + b) =
e. Recall that the binomial theorem says that
n
X
n n
a
k
k k
b =
k=0
Taking a = 1 and b =
tn =
1+
e
n!1
n!1
n
X
n (n
1) (n
k=0
2)
k!
(n
k + 1)
an
1
; we obtain
n
1
n
n
=
n
X
n (n
1) (n
k=0
1
2)
k!
(n
k + 1) 1
:
nk
k k
b :
Since n; (n
1) ; : : : ; (n
tn =
=
k + 1) are k numbers, we may rewrite this as
n
X
1 nn 1n 2
k! n n
n
k=0
n
X
1
k!
k=0
1
n
Since 1
2
n
1, 1
1
n
1
n
k+1
n
2
n
1
1
k
1
n
:
2, etc., we have
n
X
1
= sn :
k!
tn
k=0
Therefore
lim sup tn
lim sup sn = lim sn = e:
n!1
(2) lim inf n!1 tn
tn =
e. Recall that
n
X
1
k!
1
k=0
Fix m 2 N and let n
nonnegative, we have
1
k=0
1
n
Since limn!1 1
obtain
lim inf tn
n!1
=
m
X
k=0
2
n
1
1
n
1
1
k!
m
X
1
k!
2
n
1
2
n
= 1, limn!1 1
lim inf
n!1
1
n
k
1
n
1
n
1
k=0
1
k
1
n
1
2
n
1
sm for each m 2 N, we conclude that
lim inf tn
n!1
lim sm = e:
m!1
2
:
m 1
n
= 1,..., limn!1 1
= sm :
Since lim inf n!1 tn
:
m. Since the terms in the above …nite series are
m
X
1
k!
tn
n!1
n!1
k
1
n
= 1, we
Lemma (Rate of convergence of sn to e). For each n 2 N,
0<e
sn <
1
:
n!n
Proof.
e
sn =
=
1
X
1
k!
k=0
1
X
k=n+1
n
X
1
k!
k=0
1
k!
1
1
1
+
+
+
(n + 1)! (n + 2)! (n + 3)!
1
1
1
=
+
+
1+
(n + 1)!
n + 2 (n + 3) (n + 2)
!
1
1
1
+
<
1+
+
(n + 1)!
n + 1 (n + 1)2
=
=
1
1
1
(n + 1)! 1 n+1
=
1
:
n!n
Theorem 3.32. e is irrational.
Proof. Suppose that e is rational. Then there are positive integers p and q
such that
p
e= :
q
By the lemma, with n = q, we have
0<e
sq <
1
;
q!q
so that
0 < q!e
q!sq <
p
p
, we have that q!e = q! = (q
q
q
hand, by the de…nition of sq we have
Since e =
q!sq = q!
q
q
X
X
1
=
q (q
k!
k=0
1
q
1:
1)!p is an integer. On the other
1)
(k + 1)
k=0
is an integer. Therefore q!e q!sq is an integer in (0; 1), which is a contradiction.
3