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.
