(1)Problem 11649 (American Mathematical Monthly, Vol.119, June-July 2012) Proposed by Grahame Bennett (USA)

Problem 11649

(American Mathematical Monthly, Vol.119, June-July 2012) Proposed by Grahame Bennett (USA).

Let p be a real with p > 1. Let (x0, x1, . . . ) be a sequence of nonnegative real numbers. Prove that

∞

X

j=0

∞

X

k=0

x_k j + k + 1

!^p

< ∞ ⇒

∞

X

j=0

1 j + 1

j

X

k=0

xk

!^p

< ∞.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Since 2(j + 1) ≥ j + k + 1 for k = 0, . . . , j, it follows that

a_j =

∞

X

k=0

x_k j + k + 1 ≥

j

X

k=0

x_k

j + k + 1 ≥ 1 2(j + 1)

j

X

k=0

x_k =b_j 2. Hence,P∞

j=0a^p_j < ∞ implies that

∞ > 2^p

∞

X

j=0

a^p_j ≥

∞

X

j=0

b^p_j

and the proof is complete. Note that we can assume that p > 0.