Advanced book on Mathematics Olympiad

Real Analysis 495

∑

m∈M

∑∞

k= 2

∑∞

j= 1

1

mkj

=

∑

m∈M

∑∞

j= 1

∑∞

k= 2

1

mkj

.

Again, we can change the order of summation because the terms are positive. The
innermost series should be summed as a geometric series to give
∑

m∈M

∑∞

j= 1

1

mj(mj− 1 )

.

This is the same as

∑∞

n= 2

1

n(n− 1 )

=

∑∞

n= 2

(

1

n− 1

−

1

n

)

= 1 ,

as desired.

(Ch. Goldbach, solution from G.M. Fihtenholts,Kurs Differentsial’novo i Integral’no-

vo Ischisleniya(Course in Differential and Integral Calculus), Gosudarstvennoe Izda-

tel’stvo Fiziko-Matematicheskoi Literatury, Moscow 1964)

364.Let us make the convention that the letterpalways denotes a prime number. Consider

the setA(n)consisting of those positive integers that can be factored into primes that do

not exceedn. Then

∏

p≤n

(

1 +

1

p

+

1

p^2

+···

)

=

∑

m∈A(n)

1

m

.

This sum includes

∑n

m= 1

1

m, which is known to exceed lnn. Thus, after summing the

geometric series, we obtain

∏

p≤n

(

1 −

1

p

)− 1

>lnn.

For the factors of the product we use the estimate

et+t

2

≥( 1 −t)−^1 , for 0≤t≤

1

2

.

To prove this estimate, rewrite it asf(t)≥1, wheref(t)=( 1 −t)et+t

2

. Because
f′(t)=t( 1 − 2 t)et+t^2 ≥0on[ 0 ,^12 ],fis increasing; thusf(t)≥f( 0 )=1.
Returning to the problem, we have
∏

p≤n

exp

(

1

p

+

1

p^2

)

≥

∏

p≤n

(

1 −

1

p

)− 1

>lnn.