Advanced book on Mathematics Olympiad

118 3 Real Analysis

|sinn|

n

+

|sin(n+ 1 )|

n+ 1

≥

√

2 −

√

2

2

·

1

n+ 1

.

Adding up these inequalities for all odd numbersn, we obtain

∑∞

n= 1

|sinn|

n

≥

√

2 −

√

2

2

∑∞

n= 1

1

2 n

=

√

2 −

√

2

4

∑∞

n= 1

1

n

=∞.

Hence the series diverges. 

In fact, the so-called equidistribution criterion implies that iff : R → Ris a

continuous periodic function with irrational period and if

∑

n

|f (n)|

n <∞, thenf is

identically zero.

The comparison with a geometric series gives rise to d’Alembert’s ratio test:

∑∞

n= 0 an

converges if lim supn|anan+^1 |<1 and diverges if lim infn|ana+n^1 |>1. Here is a problem

of P. Erdos from the ̋ American Mathematical Monthlythat applies this test among other

things.

Example.Let(nk)k≥ 1 be a strictly increasing sequence of positive integers with the

property that

lim

k→∞

nk

n 1 n 2 ···nk− 1

=∞.

Prove that the series

∑

k≥ 1

1

nkis convergent and that its sum is an irrational number.

Solution.The relation from the statement implies in particular thatnk+ 1 ≥ 3 nkfork≥3.

By the ratio test the series

∑

k

1

nkis convergent, since the ratio of two consecutive terms

is less than or equal to^13.

By way of contradiction, suppose that the sum of the series is a rational numberpq.

Using the hypothesis we can findk≥3 such that

nj+ 1

n 1 n 2 ···nj

≥ 3 q, ifj≥k.

Let us start with the obvious equality

p(n 1 n 2 ···nk)=q(n 1 n 2 ···nk)

∑∞

j= 1

1

nj

.

From it we derive

p(n 1 n 2 ···nk)−

∑k

j= 1

qn 1 n 2 ···nk

nj

=

∑

j>k

qn 1 n 2 ···nk

nj

.