1.10. Basic inequalities 13

=

1

α

α +

1

3

α3

+

1

5

α5

+ · · · − 1

1

3

(

α2

+

α4

+ · · ·

)

=

α2

3(1 − α2)

=

1

12n(n + 1)

,

thus establishing that the sequence {an}n≥1 is decreasing and converges to

some positive number c. Therefore,

lim

n→∞

ean

= lim

n→∞

n!en

nn+

1

2

=

ec.

It remains to show that

(1.15)

ec

=

√

2π.

To do so, we call upon the well-known Wallis formula for π (proved in

1656), namely

lim

n→∞

2 · 4 · 6 · · · (2n)

1 · 3 · 5 · · · (2n − 1)

√

2n

=

π

2

,

which can be written as

(1.16)

(2nn!)2

(2n)!

1

√

2n

∼

π

2

as n → ∞.

(For an elementary proof of this formula, see W¨ astlund [145].) Using the

fact that n! ∼

nn

√

ne−nec,

we may replace n! by

nn

√

ne−nec

in (1.16), to

conclude after simplifications that

ec

∼

√

2π,

which proves (1.15).

Remark 1.15. Note that Stirling’s formula was actually proved by Abra-

ham de Moivre (1667–1754, France) and later improved by James Stirling

(1692–1770, Scotland), who established the value of the constant c. More-

over, note that it is also possible to prove the following bounds:

(1.17) 1 ≤

n!

nne−n

√

2πn

≤

e1/12n

(n ≥ 1)

(see Problem 1.15).

1.10. Basic inequalities

Two very useful inequalities are the Arithmetic Geometric Mean inequality

(AGM inequality for short) and the Cauchy-Schwarz inequality, which we

state respectively as follows.