Problem 11604
(American Mathematical Monthly, Vol.118, November 2011) Proposed by P´al P´eter D´alyay (Hungary).
Given 0 ≤ a ≤ 2, let {an}n≥1 be the sequence defined by a1 = a, an+1 = 2n−p2n(2n− an) for n ≥ 1. Find
∞
X
n=1
a2n.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let
α = 4 arcsin(p a/2) =
(arccos(2a2− 4a + 1) if a ∈ [0, 1], 2π − arccos(2a2− 4a + 1) if a ∈ [1, 2].. then, by using the fact that 2 cos2(θ/2) = 1 + cos(θ), it is easy to verify that
an = 2n−1(1 − cos(α/2n)).
Hence for N > 0
N
X
n=1
a2n =
N
X
n=1
4n−1(1 + cos2(α/2n) − 2 cos(α/2n))
=
N
X
n=1
4n−1
1 +1 + cos(α/2n−1)
2 − 2 cos(α/2n)
=1 2
N
X
n=1
4n(1 − cos(α/2n)) −1 2
N
X
n=1
4n−1 1 − cos(α/2n−1)
=1 2
N
X
n=1
4n(1 − cos(α/2n)) −1 2
N −1
X
n=0
4n(1 − cos(α/2n))
=1
2 4N(1 − cos(α/2N) − (1 − cos(α) . Finally
∞
X
n=1
a2n= 1 2
lim
N →+∞4N(1 − cos(α/2N) − (1 − cos(α))
= α2
4 + a2− 2a = 4 arcsin2(p
a/2) + a2− 2a.
