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(1)

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]. 

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