Advanced book on Mathematics Olympiad

494 Real Analysis

The problem now repeats forw 1 , which is irrational and between 0 and 1. Againp 1 has
to be the unique integer with the property that

1

p 1 + 1

< 1 −p 0 w<

1

p 1

.

If we setw 2 = 1 −p 1 w 1 , then

w=

1

p 0

−

1

p 0 p 1

+

w 2

p 0 p 1

.

Now the inductive pattern is clear. At each step we setwk+ 1 = 1 −pkwk, which is an
irrational number between 0 and 1. Then choosepk+ 1 such that

1

pk+ 1 + 1

<wk+ 1 <

1

pk+ 1

.

Note that

wk+ 1 = 1 −pkwk< 1 −pk

1

pk+ 1

=

1

pk+ 1

,

and thereforepk+ 1 ≥pk+ 1 >pk.
Once the numbersp 0 ,p 1 ,p 2 ,...have been constructed, it is important to observe
that sincewk∈( 0 , 1 )andp 0 p 1 ···pk≥(k+ 1 )!, the sequence

1

p 0

−

1

p 0 p 1

+···+(− 1 )k+^1

wk+ 1

p 1 p 2 ···pk

converges tow.Sop 0 ,p 1 ,...,pk,...have the required properties, and as seen above,
they are unique.
(13th W.L. Putnam Mathematical Competition, 1953)

363.First, denote byMthe set of positive integers greater than 1 that are not perfect
powers (i.e., are not of the forman, whereais a positive integer andn≥2). Note that
the terms of the series are positive, so we can freely permute them. The series is therefore
equal to

∑

m∈M

∑∞

k= 2

1

mk− 1

.

Expanding each term as a geometric series, we transform this into