Exponentiating, we find that
1
e n+1 ≤ 1 +
e as the limit of (1 + 1/n)n
Math 122 Calculus III
1
n
1
≤ en .
Taking the (n + 1)st power of the left inequality
gives us
n+1
e ≤ 1 + n1
D Joyce, Fall 2012
th
This is a small note to show that the number e while taking the n power of the right inequality
gives us
is equal to a limit, specifically
n
1 + n1 ≤ e.
1 n
lim 1 + n = e.
n→∞
Together, they give us these important bounds on
Sometimes this is taken to be the definition of e, but the value of e:
I’ll take e to be the base of the natural logarithms.
n
n+1
1 + n1 ≤ e ≤ 1 + n1
.
For a positive number x the natural logarithm of
x is defined as the integral
Divide the right inequality by 1 + n1 to get
Z x
1
dt.
ln x =
e
1 n
≤
1
+
1 t
n
1 + n1
Then e is the unique number such that ln e = 1,
that is,
which we combine with the left inequality to get
Z e
1
1=
dt.
e
1 n
≤ e.
1 t
1 ≤ 1+ n
1+ n
x
The natural exponential function e is the function
inverse to ln x, and all the usual properties of logae
→ e and e → e, so by the pinching
But both
rithms and exponential functions follow.
1 + n1
n
Here’s a synthetic proof that e = lim 1 + n1 . theorem
n→∞
n
1 + n1 → e,
A synthetic proof is one that begins with statements that are already proved and progresses one also.
q.e.d.
step at a time until the goal is achieved. A defect
of synthetic proofs is that they don’t explain why
Math
122
Home
Page
at
any step is made.
http://math.clarku.edu/~djoyce/ma122/
Proof. Let t be any number in an interval [1, 1 + n1 ].
Then
1
1
≤ 1.
1 ≤
t
1+ n
Therefore
Z 1+ 1
n
1
1
1+
Z
1
n
dt ≤
1
1
1+ n
1
dt ≤
t
Z
1
1+ n
1 dt.
1
1
The first integral equals n+1
, the second equals
1
ln(1 + n ), and the third equals n1 . Therefore,
1
1
1
≤
ln
1
+
≤ n.
n+1
n
1