Let {an}n≥1 be a sequence of real numbers

Problem 11809

(American Mathematical Monthly, Vol.121, December 2014) Proposed by O. Kouba (Syria).

Let {aⁿ}ⁿ≥1 be a sequence of real numbers.

(a) Suppose {aⁿ}ⁿ≥1 consists of nonnegative numbers and is nonincreasing, and P_∞

n=1aⁿ/√

n converges. Prove thatP_∞

n=1(−1)^⌊√n⌋aⁿ converges.

(b) Find a nonincreasing sequence {aⁿ}ⁿ≥1 of positive numbers such that limⁿ_→∞√naⁿ= 0 andP_∞

n=1(−1)^⌊√n⌋aⁿ diverges.

Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.

(a) We first note that limⁿ_→∞√

naⁿ= 0. Otherwise there exist ε > 0 and n0≥ 1, such that for all n ≥ n⁰,√naⁿ> ε and

n

X

k=n0

a^k

√k > ε

n

X

k=n0

1 k, which contradicts the fact thatP_∞

n=1aⁿ/√

n converges.

Moreover, we have that

Sⁿ:=

n

X

k=1

(−1)^⌊√k⌋=

a−1

X

k=1

(−1)^k((k + 1)²− k²) + (−1)^a(n + 1 − a²) = −(−1)^aa − 1 + (−1)^a(n + 1 − a²)

where a = ⌊√

n⌋. Therefore |Sⁿ| ≤ |n − a²− a| + 2 ≤ a + 2 ≤√ n + 2.

Now, for 1 ≤ n ≤ m,

m

X

k=n

(−1)^⌊√k⌋a^k =

m

X

k=n

(S^k− S^k−1)a^k=

m−1

X

k=n

S^k(a^k− a^k+1) − Sⁿ−1aⁿ+ S^ma^m.

Since {aⁿ}ⁿ≥1 is a nonincreasing sequence, we have

m−1

X

k=n

|S^k(a^k− a^k+1)| ≤

m−1

X

k=n

(√

k + 2)(a^k− a^k+1)

=

m

X

k=n

(√ k −√

k − 1)a^k+ (√

n − 1 + 2)aⁿ− (√

m + 2)a^m

≤

m

X

k=n

a^k

√k+ (√

n − 1 + 2)aⁿ− (√

m + 2)a^m.

Hence, it follows 

m

X

k=n

(−1)^⌊√k⌋ak

≤

m−1

X

k=n

|S^k(ak− a^k+1)| + |Sⁿ−1|aⁿ+ |S^m|a^m≤

m

X

k=n

a^k

√k + 2(√

n − 1 + 2)aⁿ

which implies that {Pⁿ

k=1(−1)^⌊√k⌋a^k}ⁿ≥1 is a Cauchy sequence because each term on the right hand side goes to zero as n and m go to infinity.

(b) For n ≥ 1, let

aⁿ = 1

xⁿln(1 + xⁿ) with xⁿ= ⌊(√

n + 1)/2⌋.

Then {aⁿ}ⁿ≥1 is a nonincreasing sequence of positive numbers such that for n > 1,

0 <√

naⁿ< 2√ n (√

n − 1) ln(√

n + 1)/2) because x − 1 < ⌊x⌋. Hence limⁿ→∞√

naⁿ= 0. Moreover

(−1)^⌊√n⌋= −1 and xⁿ= m for (2m − 1)²≤ n < (2m)², (−1)^⌊√n⌋= +1 and xⁿ= m for (2m)²≤ n < (2m + 1)², which implies that for N < (2m + 1)²,

N

X

n=1

(−1)^⌊√n⌋an ≥

m−1

X

n=1

2

n ln(1 + n)− (4m − 1) m ln(1 + m). Therefore the series diverges

∞

X

n=1

(−1)^⌊√n⌋aⁿ ≥

∞

X

n=1

2

n ln(1 + n) − lim_m

→∞

(4m − 1)

m ln(1 + m) = +∞.