Technical Lemmas 421

Using Stirling’s approximation we further have that

≤

(em

d

)d(

1 +

(

d

e

)d

(m−d)

(d+ 1)

√

2 πd(d/e)d

)

=

(em

d

)d(

1 +

√ m−d

2 πd(d+ 1)

)

=

(em

d

)d

·

d+ 1 + (m−d)/

√

2 πd

d+ 1

≤

(em

d

)d

·

d+ 1 + (m−d)/ 2

d+ 1

=

(em

d

)d

·

d/2 + 1 +m/ 2

d+ 1

≤

(em

d

)d

· m

d+ 1

,

where in the last inequality we used the assumption thatd≤m−2. On the
other hand,
(
em
d+ 1

)d+1

=

(em

d

)d

·

em

d+ 1·

(

d

d+ 1

)d

=

(em

d

)d

·

em

d+ 1

·

1

(1 + 1/d)d

≥

(em

d

)d

·

em

d+ 1

·

1

e

=

(em

d

)d

·

m

d+ 1,

which proves our inductive argument.

lemmaA.6 For alla∈Rwe have

ea+e−a

2 ≤e

a^2 / (^2).
Proof Observe that
ea=

∑∞

n=0

an

n!

.

Therefore,

ea+e−a

2 =

∑∞

n=0

a^2 n

(2n)!,

and

ea

(^2) / 2

∑∞

n=0

a^2 n

2 nn!

.

Observing that (2n)!≥ 2 nn! for everyn≥0 we conclude our proof.