Advanced book on Mathematics Olympiad

496 Real Analysis

Therefore,

∑

p≤n

1

p

+

∑

p≤n

1

p^2

>ln lnn.

But

∑

p≤n

1

p^2

<

∑∞

n= 2

1

k^2

=

π^2

6

− 1 < 1.

Hence

∑

p≤n

1

p

≥ln lnn− 1 ,

as desired.
(proof from I. Niven, H.S. Zuckerman, H.L. Montgomery,An Introduction to the
Theory of Numbers, Wiley, 1991)

365.We have

(k^2 + 1 )k!=(k^2 +k−k+ 1 )k!=k(k+ 1 )k!−(k− 1 )k!

=k(k+ 1 )!−(k− 1 )k!=ak+ 1 −ak,

whereak=(k− 1 )k!. The sum collapses toan+ 1 −a 1 =n(n+ 1 )!.

366.Ifζis anmth root of unity, then all terms of the series starting with themth are zero.
We are left to prove that

ζ−^1 =

m∑− 1

n= 0

ζn( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn).

Multiplying both sides byζyields the equivalent identity

1 =

m∑− 1

n= 0

ζn+^1 ( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn).

The sum telescopes as follows:

m∑− 1

n= 0

ζn+^1 ( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn)

=

m∑− 1

n= 0

( 1 −( 1 −ζn+^1 ))( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn)