Problem 12101
(American Mathematical Monthly, Vol.126, March 2019) Proposed by Hojoo Lee (South Korea).
Find the least upper bound of
∞
X
n=1
√xn+1−√xn
p(1 + xn+1)(1 + xn) over all increasing sequencesx1, x2, . . . of positive real numbers.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We will show that the least upper bound is π/2 (and it is not attained).
Let 0 < a ≤ b and let α = arctan(√
a) and β = arctan(√ b) then
√b −√
√ a 1 + a√
1 + b = (tan(β) − tan(α)) cos(α) cos(β) = sin(β − α) ≤ β − α = arctan(√
b) − arctan(√ a).
Hence, since the sequence {xn}n≥1 is positive and increasing it follows that
∞
X
n=1
√xn+1−√xn p(1 + xn+1)(1 + xn) ≤
∞
X
n=1
arctan(√xn+1) − arctan(√xn)
= arctan(√
L) − arctan(√
x1) ≤π
2 − arctan(√
x1) < π 2 where L = supn≥1{xn} ∈ (0, +∞]. So π/2 is a strict upper bound.
Let N be a positive integer and let xn = tan2 2Nnπ for n = 1, . . . N − 1 and xn = xN−1+ n − (N − 1) for all n ≥ N. Then {xn}n≥1 is positive and strictly increasing and as N → +∞
∞
X
n=1
√xn+1−√xn p(1 + xn+1)(1 + xn) >
N−2
X
n=1
√xn+1−√xn p(1 + xn+1)(1 + xn)
=
N−2
X
n=1
sin π 2N
= (N − 2) sin π 2N
→ π 2
and we may conclude that π/2 is the least upper bound.
