Problem 11505
(American Mathematical Monthly, Vol.117, May 2010) Proposed by B. Burdick (USA).
Define {an} to be the periodic sequence given by a1= a3= 1, a2= 2, a4= a6= −1, a5= −2, and an= an−6 forn ≥ 7. Let {Fn} be the Fibonacci sequence with F1= F2= 1. Show that
∞
X
k=1
akFkF2k−1
2k − 1
∞
X
n=0
(−1)kn Fkn+2k−1Fkn+3k−1
=π 4.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By d’Ocagne’s identity
FNFM+1− FN+1FM = (−1)MFN −M we have that for M = kn + 2k − 1 and N = k(n + 1) + 2k − 1
(−1)kn Fkn+2k−1Fkn+3k−1
= −Fk(n+1)+2k−1Fkn+2k+ Fk(n+1)+2kFkn+2k−1
FkFkn+2k−1Fk(n+1)+2k−1
= 1 Fk
Fk(n+1)+2k
Fk(n+1)+2k−1 − Fkn+2k
Fkn+2k−1
.
Hence
∞
X
n=0
(−1)kn Fkn+2k−1Fkn+3k−1
= 1 Fk
∞
X
n=0
Fk(n+1)+2k
Fk(n+1)+2k−1 − Fkn+2k
Fkn+2k−1
.
= 1 Fk
Φ − F2k
F2k−1
where Φ = (√
5 + 1)/2. Therefore
S :=
∞
X
k=1
akFkF2k−1
2k − 1
∞
X
n=0
(−1)kn Fkn+2k−1Fkn+3k−1
=
∞
X
k=1
ak(ΦF2k−1− F2k) 2k − 1 =
∞
X
k=1
ak(Φ)−(2k−1) 2k − 1 because FN+1− ΦFN = (−Φ)−N.
Since ak= 2Im(exp(πi/6)2k−1) then by letting z = exp(πi/6)/Φ = (√
3 + i)/(2Φ) we obtain
S = 2Im
∞
X
k=1
1 2k − 1
exp(πi/6) Φ
2k−1!
= Im
ln 1 + z 1 − z
= Arg 1 + z 1 − z
= arctan 2Im(z) 1 − |z|2
= arctan
1/Φ 1 − 1/Φ2
= arctan(1) = π 4.
