Problem 11875
(American Mathematical Monthly, Vol.122, December 2015) Proposed by D. M. B˘atinet¸u-Girugiu and N. Stanciu (Romania).
Let fn= (1 + 1/n)n((2n − 1)!! Ln)1/n. Find limn→∞(fn+1− fn) where Ln de- notes the nth Lucas number (given by L0= 2, L1= 1, and by Ln= Ln−1+Ln−2
for n ≥ 2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
It is known that,
1 + 1
n
n
= e − e
2n+ o(1/n), ((2n − 1)!!)1/n= (2n)!
2nn!
1/n
= 2n
e +ln(2) e + o(1), L1/nn =
τn+ 1
τn
1/n
= τ + o(1/n).
where τ = (1 +√
5)/2. Hence fn= 2τ n − τ(ln 2 − 1) + o(1) and it follows that
n→∞lim(fn+1− fn) = 2τ = 1 +√ 5.
