Problem 11910
(American Mathematical Monthly, Vol.123, May 2016) Proposed by C. I. Vˇalean (Romania).
LetFk be the k-th Fibonacci number. Find
∞
X
n=1
arctan 1 F4n−3
+ arctan 1 F4n−2
+ arctan 1
F4n−1 − arctan 1 F4n
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let ϕ = (√
5 + 1)/2, then for n > 0,
arctan(ϕ−(2n+1)) + arctan(ϕ−(2n−1)) = arctan ϕ−(2n+1)+ ϕ−(2n−1) 1 − ϕ−4n
= arctan
ϕ−1+ ϕ ϕ2n− ϕ−2n
= arctan
√5 ϕ2n− ϕ−2n
!
= arctan 1 F2n. Moreover, for n > 1,
arctan 1
F2n + arctan 1
F2n−1 = arctan F2n−1+ F2n−1−1 1 − F2n−1F2n−1−1
!
= arctan F2n+ F2n−1
F2nF2n−1− 1
= arctan
F2n+1
F2nF2n−1− 1
= arctan 1 F2n−2
. Hence
N
X
n=1
arctan 1
F4n−2 − arctan 1 F4n
=
2N
X
n=1
(−1)n+1arctan 1 F2n
=
2N
X
n=1
(−1)n+1
arctan(ϕ−(2n+1)) + arctan(ϕ−(2n−1))
=
2N +1
X
n=2
(−1)narctan(ϕ−(2n−1)) +
2N
X
n=1
(−1)n+1arctan(ϕ−(2n−1))
= arctan(ϕ−1) − arctan(ϕ−(4N +1))
= π
2 − arctan(ϕ) − arctan(ϕ
−(4N +1)) and
N
X
n=1
arctan 1 F4n−3
+ arctan 1 F4n−1
=
2N
X
n=1
arctan 1 F2n−1
= arctan 1 F1
+
2N
X
n=2
arctan 1
F2n−2 − arctan 1 F2n
= arctan 1 F1
+ arctan 1
F2 − arctan 1 F4N
= π
2 − arctan 1 F4N
. Finally, it follows that
∞
X
n=1
arctan 1 F4n−3
+ arctan 1 F4n−2
+ arctan 1
F4n−1 − arctan 1 F4n
= π − arctan(ϕ).
