Problem 12006
(American Mathematical Monthly, Vol.124, December 2017) Proposed by J. D. Lee (UK) and S. Wagon (USA).
Whenn is an integer and n ≥ 2, let an = ⌈n/π⌉ and bn = ⌈csc(π/n)⌉. The sequences a2, a3, . . . andb2, b3, . . . are, respectively,
1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, . . . and
1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, . . . They differ whenn = 3. Are they equal for all larger n?
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. NO, they are not equal for all n > 3.
Since sin(x) < x for x > 0, then bn ≥ an. For n ≥ 2, the strict inequality bn > an holds if there is m ∈ N+ such that n/π < m < csc(π/n) that is
n sin(π/n) < n m < π.
Since sin(x) < x −x73 for x ∈ (0, π/2], it suffices that
π − π3 7n2 < n
m < π ⇔ 0 < π − n m < π3
7n2, which is implied by (note that π2> n2/m2 and π/7 > 1/√
5) 0 < π − n
m < 1
√5 m2.
Hence, by Hurwitz’s theorem, the above inequality is satisfied by good lower Diophantine approxi- mations of the irrational number π. For example, by using the continued fraction convergents of π, we have that the inequality holds for n = 3 and m = 1 when
a3= 1 < 2 = b3, but also for n = 80143857 and m = 25510582 when
a80143857= 25510582 < 25510583 = b80143857.
