Problem 12063
(American Mathematical Monthly, Vol.125, October 2018) Proposed by H. Ohtsuka (Japan).
Letp and q be real numbers with p > 0 and q > −p2/4. Let U0= 0, U1= 1, and Un+2= pUn+1+qUn forn ≥ 0. Calculate
lim
n→∞
v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Given two real numbers p and q, then Un is the Lucas sequence of the first kind, U0= 0, U1= 1, and Un+2= pUn+1+ qUn for n ≥ 0,
whereas the Lucas sequence of the second kind is defined as
V0= 2, V1= p, and Vn+2= pVn+1+ qVn for n ≥ 0.
Such sequences satisfy the following relations that are generalizations of those between Fibonacci numbers Fn and Lucas numbers Ln (which can be obtained by letting p = q = 1):
U2n= UnVn, 2V2n = Vn2+ DUn2
where D = p2+ 4q. Here p and D are positive, and, by induction, U2n, V2n are positive too for all n ≥ 0. Now, for n ≥ 0, let
Rn =V2n+√
D + 4 U2n
2 .
Then, Rn > 0 and, by the above relations, we have that, 4Rn2 = V22n+ (D + 4)U22n+ 2√
D + 4 U2nV2n
= 2V2n+1+ 4U22n+ 2√
D + 4 U2n+1
= 4U22n+ V2n+1+ 2√
D + 4 U2n+1= 4(U22n+ Rn+1) that is Rn =pU22n+ Rn+1. Hence
R0= v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1+ Rn >
v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1.
On the other hand, for 0 < t < 1,
tR0= v u u u tt2U12+
v u u tt4U22+
s t8U42+
r
· · · + q
t2n(U22n−1+ Rn)
<
v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
t2n(U22n−1+ Rn)
<
v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1
where the last inequality holds eventually because there is some N (t) such that, for all n ≥ N(t), t2n(U22n−1+ Rn) < U22n−1.
This is due to the fact that limn→∞Vn/Un=√
D and therefore
lim
n→∞
U22n−1
U22n−1+ Rn
= 4
(√ D +√
D + 4)2 > 0 = lim
n→∞t2n. Hence, for all t ∈ (0, 1), and for all n ≥ N(t),
tR0<
v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1 < R0
which imply that
n→∞lim v u u u tU12+
v u u tU22+
s U42+
r
· · · + q
U22n−1= R0= p +pp2+ 4q + 4
2 .
Remark. By letting p = q = 1, we have that Un= Fn, the n-th Fibonacci number, and
n→∞lim v u u u tF12+
v u u tF22+
s F42+
r
· · · + q
F22n−1= 2.
