Problem 11976
(American Mathematical Monthly, Vol.124, April 2017) Proposed by R. Bosch (USA).
Given a positive real number s, consider the sequence {un}n≥1 defined by u1 = 1, u2 = s, and un+2= unun+1/n for n ≥ 1.
(a) Show that there is a constant C such that limn→∞un = ∞ when s > C and limn→∞un = 0 when s < C.
(b) Calculate limn→∞un when s = C.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let Fn= (αn−βn)/√
5 be the nth-Fibonacci number where α = (1+√
5)/2 and β = −1/α.
Then, by using the relation Fn+2= Fn+ Fn+1, it is to verify by induction that
un+1= sFn Qn
k=1kFn−k. Hence
√5 ln(un+1) =√
5 Fnln(s) −
n
X
k=1
Fn−kln(k)
!
= (αn− βn) ln(s) +
n
X
k=1
αn−kln(k) −
n
X
k=1
βn−kln(k)
= αn(ln(s) − L) + γn− βnln(s) where
L :=
∞
X
k=1
α−kln(k) ≈ 1.163451 and γn:=
∞
X
k=n+1
αn−kln(k) +
n
X
k=1
βn−kln(k)
(the series is convergent because the radius of convergence at 0 of the power seriesP∞
k=1xkln(k) is 1 and α−1∈ (0, 1)).
(a) As n goes to infinity, we have that βnln(s) → 0, and
0 ≤ |γn| αn ≤
∞
X
k=n+1
α−kln(k) + n ln(n) αn → 0.
Hence, by letting C = eL≈ 3.20096065, we have
√5 ln(un+1) = αn(ln(s/C) + o(1)) + o(1).
Therefore if s > C then ln(s/C) > 0 and limn→∞un = ∞. On the other hand, limn→∞un = 0 when s < C.
(b) If s = C then√
5 ln(un+1) = γn+ o(1) which implies that limn→∞un= ∞ because
γn≥
2n−1
X
k=n+1
αn−kln(k) +
n
X
k=1
βn−kln(k) =
n−1
X
k=1
α−kln(n + k) +
n−1
X
k=0
(−α)−kln(n − k)
= ln(n) +
n−1
X
k=1
α−k ln(n + k) + (−1)kln(n − k) ≥ ln(n).
