Advanced book on Mathematics Olympiad

Real Analysis 493

a 0 +a 1 z+a 2 z^2 +···+anzn+···

=(a 0 −a 1 )+(a 1 −a 2 )( 1 +z)+···+(an−an+ 1 )( 1 +z+···+zn)+···.

Assume that this is equal to zero. Multiplying by 1−z, we obtain

(a 0 −a 1 )( 1 −z)+(a 1 −a 2 )( 1 −z^2 )+···+(an−an+ 1 )( 1 −zn+^1 )+··· = 0.

Define the sequencebn=an−an+ 1 ,n≥0. It is positive and

∑

nbn=a^0. Because

|z|<1, the series

∑

nbnz

nconverges absolutely. This allows us in the above inequality

to split the left-hand side into two series and move one to the right to obtain

b 0 +b 1 +···+bn+··· =b 0 z+b 1 z^2 +···+bnzn+^1 +···.

Applying the triangle inequality to the expression on the right gives

|b 0 z+b 1 z^2 +···+bnzn+^1 |≤b 0 |z|+b 1 |z^2 |+···+bn|zn|+···

<b 0 +b 1 +···+bn+···,

which implies that equality cannot hold. We conclude that the sum of the series is not

equal to zero.

362.If such a sequence exists, then the numbers

1

p 0 p 1

−

1

p 0 p 1 p 2

+

1

p 0 p 1 p 2 p 3

−··· and

1

p 0 p 1 p 2

−

1

p 0 p 1 p 2 p 3

+···

should both be positive. It follows that

0 <

1

p 0

−w=

1

p 0 p 1

−

1

p 0 p 1 p 2

+

1

p 0 p 1 p 2 p 3

−···<

1

p 0 p 1

<

1

p 0 (p 0 + 1 )

.

Hencep 0 has to be the unique integer with the property that

1

p 0 + 1

<w<

1

p 0

.

This integer satisfies the double inequality

p 0 <

1

w

<p 0 + 1 ,

which is equivalent to 0< 1 −p 0 w<w.

Letw 1 = 1 −p 0 w. Then

w=

1

p 0

−

w 1

p 0

.