Problem 12118
(American Mathematical Monthly, Vol.126, June-July 2019) Proposed by H. Ohtsuka (Japan).
Compute
∞
X
n=0
1 F2mn+ iFm
wherem is an odd integer and Fn is then-th Fibonacci number.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By using the known identity
Fi+jFi+k− FiFi+j+k = (−1)iFjFk
with i, j, k ∈ Z, it follows that for any odd integer m, Fm= F−mand F2nm+mF2nm−m− F2nm2 = Fm2 =⇒ F(2n+1)mF(2n−1)m= F2nm2 + Fm2 FmF2nm−m− F−mF2nm+m= −F2mF2nm =⇒ F2nm
F(2n+1)mF(2n−1)m
= Fm/F2m
F(2n−1)m− Fm/F2m
F(2n+1)m
F2nm+mF2nm−2m− F2nmF2nm−m= −FmF2m =⇒ Fm
F(2n+1)mF(2n−1)m
=F2nm/F2m
F(2n+1)m −F(2n−2)m/F2m
F(2n−1)m
.
Therefore, 1 F2mn+ iFm
=F2nm− iFm
F2nm2 + Fm2 = F2mn− iFm F(2n+1)mF(2n−1)m
= Fm
F2m
1
F(2n−1)m − 1 F(2n+1)m
− i F2m
F2nm
F(2n+1)m −F(2n−2)m
F(2n−1)m
Hence the given sum is telescopic:
∞
X
n=0
1 F2mn+ iFm
= Fm
F2m
∞
X
n=0
1
F(2n−1)m − 1 F(2n+1)m
− i F2m
∞
X
n=0
F2nm
F(2n+1)m −F(2n−2)m
F(2n−1)m
= Fm
F2mF−m− i F2m
φ−m−F−2m
F−m
= 1 F2m − i
F2m
φ−m+F2m
Fm
= 1 − iφm F2m
where φ = (1 +√
5)/2 and by the Binet’s formula
n→∞lim F2nm
F(2n+1)m
= φ−m and F2m
Fm
= φm− φ−m.
