Prove that for any sequence of positive numbers (a n ) n ≥1 we have

2.5.6. Prove that for any sequence of positive numbers (a n ) n ≥1 we have

2.5 Qualitative Results 93

n =1 a 1 +a 2 + ··· + a n

n =1 a n Putnam Competition, 1964

Solution. We first observe that the Cauchy–Schwarz inequality implies

(1 + 2 + ···+ n) 2 ≤ (a 1 +a 2 + ··· + a n )

a 1 a 2 a n Hence 2n +1

2n +1

a 1 ≤4 + ··· + a n n 2 (n + 1) 2 k ∑ =1 a k

2n ∑ +1 ≤4 ∑ 2 2 ∑ =4 ∑ ∑ 2 2 .

. It follows that

∑ n 2 (n + 1) 2 = ∑ n 2 −

and the conclusion follows. ⊓ ⊔

be a convergent series of real numbers. Does the series ∑ n =1 a n |a n | converge, too?

2.5.7. Let ∞ ∑

n =1 a n

√ Solution. For any integer n √ ≥ 0, define a 3n +1 =a 3n +2 = 1/ n + 1 and a 3n +3 =

−2/ n + 1. Then the series ∑ ∞

n =1 a n converges, but ∑ n =1 a n |a n | diverges. ⊓ ⊔

2.5.8. Let (a n ) n ≥1

be a sequence of real numbers such that 0 <a n ≤a 2n +a 2n +1 , for all integers n ≥1 . Prove that the series ∑ ∞ n =1 a n diverges.

Putnam Competition, 1994 Solution. We argue by contradiction and assume that the series converges. Then

∑ a n ≤ ∑ (a 2n +a 2n +1 )= ∑ a n < ∑ a n ,

contradiction. ⊓ ⊔ An immediate consequence of the following problem is that both the harmonic

series ∞ ∑ n =1 (n) −1 and the Riemann series ∞ ∑ n =2 (ln n) −1 are divergent.

2.5.9. Let (a n ) n ≥1

be a sequence of positive numbers such that a n <a n +1 +a n 2 ,

for all n ≥1 . Prove that the series ∞ ∑ n =1 a n is divergent.

94 2 Series Solution. We have: a 2 <a 3 +a 4 <a 4 +a 9 +a 5 +a 16 < ··· . Let I k denote the

set of positive subscripts that appear at the kth step in the above iteration, that is,

I 1 = {2}, I 2 = {3,4}, I 3 = {4,5,9,16}, etc. We observe that the length of I k is less than or equal to 2 k −1 . Moreover, if

k |=2 −1 for all k ∞ ≥ 1, then ∑ p ∈I k a p >a 2 , which implies that all remainders of ∑ n =1 a n are greater than a 2 , and so the series ∑ ∞

|I k

n =1 a n diverges. In order to prove that

|I k

k |=2 −1 for all k ≥ 1, we argue by contradiction and denote by m the least positive integer m such that i +1=j 2 , for some i ,j∈I m . We

observe that the numbers j 2 − 1, j 2 − 2,..., j 2 − 2 j + 2 are not squares. On the other hand, I = (I

+ 1) ∪I 2 , where I

+ 1 := {q+1; q ∈ I } and I k 2 k −1 k −1 k −1 k −1 k −1 = {q 2 ;q ∈

k −1 }. For all k < m, the definition of m implies that (I k −1 + 1) ∩ I k −1 = /0. Therefore j 2 −2∈I m −1 ,j 2 −3∈I m −2 , . . ., j 2 −2j+2∈I m −2 j+3 . So, m − 2 j + 3 ≥ 1, that is, m ≥ 2 j − 2. Since, obviously, j ≥ m + 1 for all j ∈ I m , we obtain m ≥2j−2≥

2 (m + 1) − 2 = 2m, contradiction. ⊓ ⊔

n ) n ≥1 and a sequence of nonzero real numbers ( α n ) n ≥1 such that the series ∑

2.5.10. (a) Find a sequence of distinct complex numbers (z

n =1 α n |z − z n | either con- verges to a positive number or diverges to +∞ for almost all complex numbers z ,

but not all α n are positive. (b) Let

be a sequence of distinct complex numbers. Assume that ∑ n =1 α n is an absolutely convergent series of real numbers such that ∑ ∞ n =1 α n |z − z n | −1 converges to a nonnegative number, for almost all z ∈C . Prove that α n are nonnegative for all n ≥1 .

(z ∞ n ) n ≥1

T.-L. R˘adulescu and V. R˘adulescu, Amer. Math. Monthly, Problem 11304 Solution. (a) We prove that the series

1 + ··· (2.9) |z|

diverges to

1 +∞ for all z ∈ C \ 1

2 , ··· ,± n ,...

2 n ,... series has the same nature as the series −1 + 1 + 1 + 1 + ···, which diverges. Next,

Indeed, we first observe that for any fixed z ∈C\

we observe that

≥ 0 if and only if Rez ≤ − |z|

2n and

≥ 0 if and only if Rez ≥ |z|

2n The above relations show that for any z ∈C\

2 ,...,± 1 n ,...

|z| +∑ k =1

there exists N ∈ N such that − 1 N

1 > 0. It remains to prove

that this is also true if z = iy, y ∈ R \ {0}. For this purpose

2.5 Qualitative Results 95 we observe that

provided 2 |y| ≥ (4n 2 − 8n + 3) −1 . In conclusion, the series (2.9) diverges to +∞ for all z

2 ,...,± n ,... Another example of series with the above properties is

1 ∈C\ 1

|z| ∑ n =2 n |z + lnn| |z − lnn|

, z ∈ C \ {0,±ln2;±ln3,...}.

(b) It is sufficient to focus on an arbitrary term of the sequence, say α 1 , and to show that α 1 ≥ 0. We can assume, without loss of generality, that z 1 = 0. Fix arbitrarily ε

∈ (0,1). Since ∑ n =1 | α n | < ∞, there exists a positive integer N such that ∑ i =N+1 | α i |< ε . Next, we choose r > 0 small enough that |a i | > r/ ε , for

all i ∈ {2,...,N}. Set

f (z) = ∑

n =1 |z − z n |

It follows that "

B r (0) |z| ∑

f (z)dz = α 1 + α i

B r (0) |z − z i | i =N+1

B r (0) |z − z i | i =N+1

B r (0) |z − z i |

=2 π r α 1 + ∑ | α i |

B r (0) |z − z i | i =N+1

B r (0) |z − z i |

≤2 π r α 1 + ∑ | α i |

dz

+ ε sup

dz

i ≥N+1 B r (0) |z − z i | For every i ∈ {2,...,N} we have |z − z i | ≥ |z i | − |z| ≥ r ε − r = r(1 − ε )/ ε , so

επ r

dz =

1 − ε If i ≥ N +1 we distinguish two cases: either |z i | ≥ 2r or |z i | < 2r. In the first situation

B r (0) |z − z i |

r (1 − ε ) B r (0)

we deduce that |z − z i | ≥ r, for any z ∈ B r (0). Thus "

dz

dz = π r .

B r (0) |z − z i |

r B r (0)

96 2 Series If |z i | < 2r then

dz

dz

=8 π r .

B r (0) |z − z i |

B 4r (z i ) |z − z i |

The above two relations show that

dz

sup

≤8 π r .

i ≥N+1 B r (0) |z − z i |

Using (2.10), (2.11), and (2.12), we obtain

0 ≤2 π r α 1 + ∑ | α i |·

επ r

+8 επ r .

i =2

Dividing by r and letting ε →0 we deduce that α 1 ≥ 0. ⊓ ⊔ Open problem. For part (a) of this problem, we have not been able to find an example of a series ∑ ∞

n =1 α n |z−z n | that converges to a positive number for almost all complex numbers z, but with not all α n positive. It might be possible that such

a series does not exist and the unique situation that can occur is that under our assumptions described in (a), the series ∑ ∞ =1 α n |z − z n | −1 n always diverges to +∞.

2.5.11. Suppose that a 1 >a 2 > ··· and lim n → ∞ a n =0 . Define

j ∑ −n (−1) a j =a n −a

n +1 +a n +2 − ··· .

j =n

Prove that the series ∑ ∞

n =1 S 2 n , ∑ ∞ n =1 a n S n , and ∑ n =1 a 2 n converge or diverge together. W. Trench, Amer. Math. Monthly, Problem 10624

Solution. By the Leibniz alternating series test, S n exists and satisfies 0 <S n <

a n . Thus

n =1 n implies finiteness of ∑ =1 a . To prove it, we use S n =a n

So it suffices to show that finiteness of ∞ ∑ S 2 ∞

−S n ≤ 2(x +y 2 ) to infer

+1 and the inequality (x + y) 2 2 n

∑ (S

n +S n +1 ) ≤ ∑ 2 (S n +S n +1 )=2 ∑ S n +2 ∑ S n +1 <4 ∑ S n .

Using Stirling’s formula

lim

n → ∞ n n e −n √ 2 π n

we deduce that if a n = (2n)!/4 n (n!) 2 , then a n = O(1/ n ) as n→∞. On the other hand, a n +1 /a n = (2n + 1)/(2n + 2) < 1, so (a n ) n ≥1 is a decreasing sequence of

2.5 Qualitative Results 97 positive numbers converging to 0. Thus, by the Leibniz alternating series test, the

series ∞ ∑ n =1 (−1) n a n is convergent. At an elementary level, however, the convergence of the series ∞ ∑

n =1 (−1) a n may be a little more difficult to obtain, the hard part being to show that a n →0 as n→∞. For this purpose one may use the following simple

result. ⊓ ⊔

2.5.12. Suppose (a n ) n ≥1 is a decreasing sequence of positive numbers and for each natural number n , define b n =1−a n +1 /a n . Then the sequence (a n )

n ≥1

converges to zero if and only if the series ∑ n =1 b n diverges. Solution. We first observe that for M > N,

∑ n −a n +1 )≤ ∑ b n ≤ ∑ (a n −a n +1 )=

a M Now, if ∞ ∑ n =1 b n converges and a

n →0, then letting M→∞ gives 1 ≤ ∑ n =N b n , which is impossible. Conversely, if a n → α > 0 then let N = 1 and let M→∞. The inequality above yields ∞ ∑ n =1 b n ≤ (a 1 − α )/ α , and so the series converges.

Returning now to the above example, we see that b n = 1/(2n + 2) for any n

≥ 1, so the divergence of the series ∑ ∞ n =1 b n implies that a n →0 as n→∞. The same technique gives an easy proof of the convergence of such series as

∞ (−1) n n n

n =1

and the series of binomial coefficients

with α

n =[α]+1

be a convergent series of positive numbers and set r n = ∑ ∞

2.5.13. Let ∞ ∑ n =1 a n

k =n a k . Prove that the series ∞ ∑ n =1 a n /r n diverges.

O. Kellogg Solution. By definition, r n =a n +r n +1 . Hence

r n +1 By induction we establish that for all positive numbers n and p we have

r n +p Since the series ∑ ∞ n =1 a n is convergent, we deduce that r n +p →0 as p→∞. Fix δ ∈

r n +1

(0, 1). Hence, to any positive integer n, there corresponds p ∈ N such that

a n a n +1

a n +p

r n +p

r n +1

r n +p

98 2 Series

It follows that the series ∑ ∞ n =1 a n /r n diverges.

be an arbitrary series. Each increasing sequence n 1 ,n 2 ,n 3 , . . . of positive integers determines a series

Let ∞ ∑ n =1 a n

a n 1 +a n 2 +a n 3 + ··· ,

which is called a subseries of the given series. The following exercise establishes an interesting property of some subseries of a divergent series. ⊓ ⊔

2.5.14. Let (a ∞ n ) n ≥1

be a sequence of positive numbers such that lim n → ∞ a n =0 and the series ∑ n =1 a n diverges. Let a be a positive number. Then there is a subseries

of ∞ ∑ n =1 a n that converges to a .

Solution. Let n 1 be the least integer such that a n < a/2 for all n ≥ n 1 . Let m 1 be the greatest integer for which

s 1 =a n 1 +a n 1 +1 + ··· + a m 1 < a. Then a /2 < s 1 < a. Let n 2 be the least integer such that n 2 >m 1 and a n < (a −s 1 )/2,

provided n ≥n 2 . Let m 2 denote the greatest integer for which s 2 =s 1 +a n 2 +a n 2 +1 + ··· + a m 2 < a. Then a

− a/2 2 <s

2 < a. Let n 3 be the least integer such that n 3 >m 2 and a n < (a − s 2 )/2, provided n ≥ n 3 . Let m 3 denote the greatest integer for which

s 3 =s 2 +a n 3 +a n 3 +1 + ··· + a m 3 < a. Then a − a/2 3 <s 3 < a. Continuation of the construction yields the sequences of

integers (n k ) k ≥1 and (m k ) k ≥1 such that

n 1 <m 1 <n 2 <m 2 <n 3 <m 3 < ··· , s q =s q −1 +a n q +a n q +1 + ··· + a m q for all q ≥ 2,

and a − a/2 q <s q < a. It follows that the required subseries converging to a is

a n 1 +a n 1 +1 + ··· + a m 1 +a n 2 +a n 2 +1 + ··· + a m 2 +a n 3 +a n 3 +1 + ··· . ⊓ ⊔

2.5.15. Let ( λ n ) n ≥1

be a sequence of positive numbers. Set a 1 =1 , a n +1 =

a n + λ n a −1 n for n ≥1 . Prove that lim n → ∞ a n exists if and only if the series ∑ ∞ n =1 λ n converges.

T. Davison Solution. We first observe that a n +1 >a n ≥ 1 for all n ≥ 1. Assume that

lim n → ∞ a n = a. Hence, for all n, we have a n ≤ a. Therefore

∑ (a n +1 −a n )=a N +1 −a 1 = ∑ ≥ ∑ λ n .

n =1

n =1 a n

a n =1

2.5 Qualitative Results 99

It follows that ∑ ∞ n =1 λ n ≤ a(a − 1) < ∞. Conversely, assume that the series ∞ ∑ n =1 λ n converges. Then

∑ (a n +1 −a n )=a N +1 −a 1 = ∑ a n ≥ ∑ λ n .

Therefore lim

n → ∞ a n =1+∑ n =1 λ n . Since the harmonic series ∞ ∑ n =1 1 /n is divergent, the partial sums S n =∑ n k =1 1 /k are larger than every real number, provided n is big enough. A remarkable property

is that S n is never an integer, for all n ≥ 2. Another striking result related to the harmonic series is the following. ⊓ ⊔