Problem 12090
(American Mathematical Monthly, Vol.126, February 2019) Proposed by H. Ohtsuka (Japan).
The Pell-Lucas numbers Qn satisfy Q0= 2, Q1= 2, and Qn= 2Qn−1+ Qn−2 for n ≥ 2. Prove
∞
X
n=1
arctan
2 Qn
arctan
2 Qn+1
= π2 32.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We first note that Qn= qn+ (−q)−nwith q = 1 +√ 2.
Therefore, if n is even
arctan
2 Qn
= arctan
2
qn+ q−n
= arctan
q−(n−1)− q−(n+1) 1 + q−(n−1)· q−(n+1)
= arctan(q−(n−1)) − arctan(q−(n+1)).
On the other hand, if n is odd
arctan
2 Qn
= arctan
2
qn− q−n
= arctan
2q−n 1 − q−2n
= 2 arctan(q−n).
Hence the desired sum is
S=
∞
X
n=1
arctan
2
Q2n−1
arctan
2 Q2n
+
∞
X
n=1
arctan
2 Q2n
arctan
2
Q2n+1
=
∞
X
n=1
arctan
2 Q2n
arctan
2
Q2n−1
+ arctan
2
Q2n+1
= 2
∞
X
n=1
arctan(q−(2n−1)) − arctan(q−(2n+1))
·
arctan(q−(2n−1)) + arctan(q−(2n+1))
= 2
∞
X
n=1
arctan2(q−(2n−1)) − arctan2(q−(2n+1))
= 2 arctan2(q−1) − 2 lim
n→∞arctan2(q−(2n+1))
= 2 arctan2(√
2 − 1) = 2π 8
2
=π2 32.
