Does there exist a bijective map π :N →N such that

2.5.18. Does there exist a bijective map π :N →N such that

∞ π (n)

n =1 n

International Mathematical Competition for University Students, 1999 Solution. We argue that the series ∑ ∞ n =1 π (n)/n 2 is divergent for any bijective

map π :N → N. Indeed, let π

be a permutation of N. For any n ∈ N, the numbers π (1), . . . , π (n) are distinct positive integers; hence

n (n + 1) π (1) + ··· + π (n) ≥ 1 + ··· + n =

2 Therefore

π (n) ∞ 1

∑ 2 = ∑ ( π (1) + ··· + π (n)) 2 −

n =1 n

(n + 1) 2

n =1

∞ n (n + 1)

2n +1

2n +1

n =1

2 n (n + 1)

n =1 2n (n + 1) n =1 n +1

2.5 Qualitative Results 101 An alternative argument is based on the observation that for any positive integer

N we have

3N

π (n) 1

n =N+1

Indeed, only N of the 2N integers π (N + 1), . . . , π (3N) are at most N, so that at least N of them are strictly greater than N. Hence

3N

π (n)

1 3N

=N+1 ∑ n 2 ≥ (3N) 2 π n (n) > =N+1 ∑ 9N 2 ·N·N= .⊓ n ⊔ 9

2.5.19. Let a 0 =b 0 =1 . For each n ≥1 let

Show that a n ·b n is an integer. Solution. We prove by induction on n that a n /e and b n e are integers, for any

n ≥ 1. From the power series of e x , we deduce that a 1 =e 1 = e and b n =e −1 = 1/e. Suppose that for some n ≥ 1, a 0 ,a 1 ,...,a n (resp., b 0 ,b 1 ,...,b n ) are all integer multiples of e (resp., 1 /e). Then, by the Newton’s binomial,

∞ (k + 1) n +1

∞ (k + 1) n

a n +1 = ∑

k =0 (k + 1)!

k =0

m =0 ∑ k =0 k !

∑ k (−1) =−

k +1

m =0 m

The numbers a n +1 and b n +1 are expressed as linear combinations of the previous elements with integer coefficients, which concludes the proof. ⊓ ⊔

102 2 Series The following result is closely related to the divergence of the harmonic series

and establishes an interesting property that holds for all sequences of positive real numbers.

2.5.20. Let (a n ) n ≥1

be a sequence of positive numbers. Show that

Solution. Arguing by contradiction, we deduce that there exists a positive integer N such that for all n ≥ N,

1 +a n

n +1

This inequality may be rewritten as

Summing these inequalities, we obtain

N which is impossible, since the harmonic series is divergent. ⊓ ⊔

n +1

The following result is optimal, in the sense that (under the same hypotheses) there is no function f with f (n)/n→∞ and f (n)a n →0 as n→∞.

2.5.21. Let (a n )

be a sequence of real numbers such that the series ∑ n =1 a n converges and a n ≥a n +1 ≥0 for all n ≥1 . Prove that na n →0 as n →∞ .

∞ n ≥1

Solution. Assume that m and n are positive integers with 2m < n. Since the series ∑ ∞ n =1 a n converges, we may assume that r m : =∑ ∞ j =m+1 a j < ε , where ε > 0 is fixed

arbitrarily. Then

(n − m)a n ≤a m +1 + ··· + a n ≤r m : = ∑ a j .

j =m+1

Thus,

0 < na n ≤

n −m

n −m

which concludes the proof. ⊔ ⊓

be a sequence of positive numbers such that the series ∑ n =1 a n converges. Prove that there exists a nondecreasing divergent sequence (b n ) n ≥1 of positive integers such that the series ∞ ∑ n =1 a n b n also converges.

2.5.22. Let (a ∞ n ) n ≥1

Solution. Since ∑ ∞ n =1 a n converges, we deduce that for any n ≥ 1, there exists a positive integer N (n) depending on n such that

2.5 Qualitative Results 103

j =N(n)

To conclude the proof, it is enough to take b m = n, provided N(n) ≤ m < N (n + 1). ⊓ ⊔

“Prime numbers were invented to multiply them.” Israel Gelfand attributes this famous quotation to the Russian physicist Lev Landau (1908–1968), who was awarded the 1962 Nobel Prize in Physics. Nonetheless, the additive properties of the prime numbers have been a fascinating subject for generations of mathemati- cians, and by now there is an ample supply of powerful and flexible methods to attack classical problems.

Problems concerning infinite sums of positive integers play an important role in number theory. For instance, the Hungarian mathematicians Paul Erd˝os (1913–1996) and Paul Tur´an (1910–1976) conjectured in 1936 that every infinite

set {a 1 ,a 2 , . . .} of positive integers contains arbitrarily long arithmetic progressions, provided that the series ∑ ∞ n =1 1 /a n diverges. This conjecture is still open, and it is not even known whether such a set contains an arithmetic progression of length 3.A related theorem due to the Dutch mathematician Bartel Leendert van der Waerden

(1903–1996) states that for any partition N + = n j =1 S j into finitely many subsets S j , at least one of them contains arbitrarily long arithmetic progressions. A remark-

able breakthrough in the field was achieved in 2004 by the British mathematician Ben Green and Fields Medalist Terence Tao [35], who proved a long-standing con- jecture, namely, that the set of prime numbers contains arbitrarily long arithmetic progressions. For example, 7, 37, 67, 97, 127, 157 is an arithmetic progression of length 6 consisting only of primes. In 2004, Frind, Underwood, and Jobling found a progression of length 23 with first element of size

≈ 5.6 × 10 13 . Notice that the Green–Tao theorem guarantees arbitrarily long arithmetic progressions of prime

numbers, but infinitely long such progressions do not exist: n + kd is not prime if k = n.

The following result establishes that the series of inverses of prime numbers diverges. We provide three different proofs to this property. It may be surprising to note that the sum of the reciprocals of the twin prime numbers (that is, pairs of primes that differ by 2, such as 11, 13 or 17, 19) is a convergent series, see [65, pp. 94–103].