Prove that if is a sequence of positive numbers with (a ∞ n ) n ≥1 ∑ n =1 a n <∞ , then for all p ∈ (0,1) ,

2.5.28. Prove that if is a sequence of positive numbers with (a ∞ n ) n ≥1 ∑ n =1 a n <∞ , then for all p ∈ (0,1) ,

lim n 1 −1/p p (a

1 + ··· + a n )

p 1 /p

=0. Grahame Bennet

n →∞

Solution. (Eugene A. Herman) Define the sequence (x

n ) n ≥1 by x n = (1/n) 1 −p (a 1 + ··· + a n ). Since

pp

1 + ··· + a n )) =x n , it suffices to show that x n → 0 as n→∞. Using H¨older’s inequality we obtain

n 1 −1/p p (a

1 + ··· + a n

p ) 1 /p = ((1/n) 1 −p p (a

Given any ε > 0 choose N such that ∑ ∞ n =N a /p

n <( ε /2) 1 . For any n larger than both

pp

N and ((a 1 + ··· + a N )2/ ε ) 1 /(1−p) , we have

k =N+1

1 ε p n −p n p

1 −p n

2 k =N+1 n

k =N+1

k =N+1

2 2 which concludes the proof. ⊓ ⊔

2.5.29. Let (a n ) n ≥1

be a sequence of real numbers such that ∞ ∑ n =1 |a n | is conver- gent and ∞ ∑ n =1 a kn =0 for each positive integer k . Prove that a n =0 for all n ≥1 .

108 2 Series Solution. Let p n

be the set of positive integers that are not divisible by any p 1 ,...,p n . Note that the first two members of R n are 1 and p n +1 . We first claim that

be the nth prime greater than 1, and let R n

∑ a j =0

j ∈R n

for each n. From this it follows that

|a ∞

1 |≤∑ p n +1 |a j | for all n, and therefore that

a 1 = 0. A similar argument shows that for any k, ∑ j ∈R n a kj = 0, and hence that |a k | = 0.

For a bounded sequence b = (b n ) n ≥1 , write n b n , where a = (a n ) n ≥1 . Let c k

be the sequence with 1 in place nk (n = 1, 2, . . .) and 0 elsewhere. The hypothesis states that

Fix an integer n, and for 1 ≤ r ≤ n, let T r

be the set of products of r distinct primes chosen from p 1 ,...,p n . Consider the sequence b defined by

b =c 1 + ∑ (−1) r k

r =1

k ∈T r

If we show that b j is 1 for j + ∈R n in R and 0 for other j, then our claim follows. If j is n , then c k j = 0 for all k in n r =1 T r , so b j =c 1 j = 1. Suppose now that j is not in

R n , and let q 1 ,...,q m

be the primes that divide j and are not greater than p n . Let U r

be the set of products of r distinct primes chosen from q 1 ,...,q m . Then U r has C r m members, and c k j = (−1) r for each k in U r . These, together with c 1 , are the only c k that make a nonzero contribution to b j , which is therefore equal to

1 + ∑ (−1) r C r m = (1 − 1) m = 0.

r =1

This completes the proof. ⊓ ⊔ We say that a function f : R

→R is convergence-preserving if for every conver- gent series

∑ ∞ n =1 a n , the series ∑ n =1 f (a n ) also converges. For example, all functions

f (x) = ax are convergence-preserving, where a ∈ R is an arbitrary constant. The fol- lowing properties are also true:

(i) if f is convergence-preserving, then f (0) = 0; (ii) if f and g are convergence-preserving, so are f + g, a f , f ◦ g, and g(x) = f (ax), where a is a constant.

The following exercise establishes that every convergence-preserving function has at most linear growth in a neighborhood of the origin.

2.5.30. Let f :R →R

be a convergence-preserving function. Prove that there exist

a real number M and ε >0 such that f (x) < Mx for all 0 <x< ε. Solution. Arguing by contradiction, we deduce that for any positive integer n we

can find 0 <x n < 1/n 2 such that f (x n ) > nx n . Let j n

be the least integer such that j n ≥ 1/(n 2 x n ). Then j n x n < 2/n 2 . Consider the series

2.5 Qualitative Results 109 x 1 +x 1 + ··· + x 1 (j 1 times )+x 2 +x 2 + ··· + x 2 (j 2 times ) + ··· .

This series converges by comparison with ∑ n =1 2 /n 2 .

When we apply f to each term of the series and sum the first n blocks, we obtain

∑ j n f (x n )≥ ∑ nj n x n ≥ ∑ ,

which diverges, contradicting the hypothesis that f is convergence-preserving. As proved in [116], a function f is convergence-preserving if and only if there

exists a constant a ∈ R such that f (x) = ax in a neighborhood of the origin. This result guarantees the existence of many series having strange properties. For ex- ample, there is a sequence (a n )

n =1 a n diverges but ∑ n =1 tan a n con- verges; otherwise, (arctan x) 3 would be convergence-preserving. Similarly, there is

for which ∑

n =1 n convergent and ∑ n =1 tan b n divergent; otherwise, tan 3 x would be convergence-preserving. ⊓ ⊔

a sequence (b n ) n with ∞ ∑ b 3 ∞

be a convergent series with sum A and with positive terms that satisfy a n +1 ≤a n ≤ 2a n +1 for all n ≥1 . Prove that any positive number B <A is the sum either of a finite number of terms or of a subseries of the considered series.

2.5.31. Let ∞ ∑ n =1 a n

Marian Tetiva, Problem 101, GMA 1(2001) Solution. We first assume that a 1 ≤ B < A. Define A n =∑ n k =1 a k . Then

hence there exists s ≥ 1 for which B ∈ [A s ,A s +1 ). If B = A s , we are done. Otherwise, we have B ∈ (A s ,A s +1 ), and thus

0 <B−A s <a s +1 . Since a 1 ≥a 2 ≥ ··· and lim n →∞ a n = 0, we have

n ≥s+1

hence there exists p ∈N ∗ such that B −A s ∈ [a s +p+1 ,a s +p ). Again, we are done if

B −A s =a s +p+1 . If not, according to the hypothesis,

0 <B−A s −a s +p+1 <a s +p −a s +p+1 ≤a s +p+1 .

In turn, B −A s −a s +p+1 is in an interval of the form [a s +p+q+1 ,a s +p+q ) for some positive integer q. Therefore we obtain

0 ≤B−A s −a s +p+1 −a s +p+q+1 <a s +p+q −a s +p+q+1 ≤a s +p+q+1 .

110 2 Series So we can assume in general that we have natural numbers s <t 1 < ··· < t k

such that

0 ≤B−A s −a t 1 − ··· − a t k <a t k −1 −a t k ≤a t k . We are done if we have equality in the left-hand inequality; otherwise, B −A s −

a t 1 − ··· − a t k is in an interval [a t k +u+1 ,a t k +u ) (for some natural u), and with t k +1 = t k + u + 1 we have

0 ≤B−A s −a t 1 − ··· − a t k −a t k +1 <a t k +1 −1 −a t k +1 ≤a t k +1

and so on. This method yields a strictly increasing sequence of natural numbers s <t 1 <t 2 < ··· with the property that

0 ≤B−A s −a t 1 − ··· − a t n <a t n −1 −a t n ≤a t n , ∀n ∈ N ∗ . If any of the left-hand inequalities becomes an equality, the problem is solved, since

obviously, B is the sum of a finite number of terms of the series ∑ ∞ n =1 a n . On the other hand, if all these inequalities are strict, because of lim n →∞ t n = ∞ we also have lim n →∞ a t n = 0 and

B = lim n (A

→∞ s +a t 1 + ··· + a t n ),

which proves the claim. In the case 0 <B<a

1 , notice that a 1 ≤ A − B < A. Indeed, 2A − 2a 1 = ∑ n =1 2a n ∞ +1 ≥∑ n =1 a n = A, whence A − a 1 ≥a 1 > B (the other case is clear). In conformity with the part above, there is a sequence j 1 <j 2 < ··· of natu- ral numbers such that A

n =1 a j ∞ n , which easily implies B =∑ n =1 a i n , for {i 1 ,i 2 ,...}=N ∗ \{j 1 ,j 2 , . . . }. ⊓ ⊔

−B=∑ ∞