Problem 12153
(American Mathematical Monthly, Vol.127, January 2020) Proposed by O. Kouba (Syria).
For a real number x whose fractional part is not 1/2, let hxi denote the nearest integer to x. For a positive integer n, let
an=
n
X
k=1
1 h√
ki − 2√ n.
(a) Prove that the sequence (an)n≥1is convergent, and find its limit L.
(b) Prove that the set {√
n(an− L) : n ≥ 1} is a dense subset of [0, 1/4].
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. It is easy to verify that
h√ ki =
√ k +1
2
= j for k ∈ [j(j − 1) + 1, j(j + 1)],
and therefore
an+ 1 =
n
X
k=1
1 h√
ki− 2√
n + 1 = n h√
ni+ h√
ni − 1 − 2√
n + 1 = (√ n − h√
ni)2 h√
ni .
(a) As n goes to infinity, we have that h√
ni → +∞ and
0 ≤ an+ 1 = (√ n − h√
ni)2 h√
ni = {√
n +12} − 12
2 h√
ni ≤ 1/4
h√ ni → 0 where {x} = x − ⌊x⌋ ∈ [0, 1), and we may conclude that an→ L = −1.
(b) We have that
√n(an− L) =√
n(an+ 1) =
√n(√ n − h√
ni)2 h√
ni .
Let j = h√
ni then n ∈ [j(j − 1) + 1, j(j + 1)] and
0 ≤√
n(an+ 1) ≤ pj(j + 1)(pj(j + 1) − j)2
j = 1
4 · pj(j + 1)
√ j+1+√
j 2
2 < 1/4
where at the last step we applied the AM-GM inequality.
Let α ∈ [0, 1/4]. Then there exist two sequences of integers (pk)k≥1 and (qk)k≥1 with 0 ≤ pk ≤ qk and qk → +∞ such that pk/qk→ 2√
α ∈ [0, 1]. Let nk= q2k+ pk then
q2k≤ nk≤ q2k+ qk <
qk+1
2
2
=⇒ h√nki = qk.
Therefore, as n goes to infinity,
√nk− h√
nki =√
nk− qk= nk− qk2
√nk+ qk
= pk
pqk2+ pk+ qk
→√ α
and √nk(ank+ 1) =
√nk(√nk− h√nki)2 h√nki → α which means that the set {√
n(an+ 1) : n ≥ 1} is a dense subset of [0, 1/4].
