Problem 12129
(American Mathematical Monthly, Vol.126, August-September 2019) Proposed by H. Ohtsuka (Japan).
Compute
v u u t2 +
s 2 +
r
2 + · · · + q
2 −√ 2 + · · ·
where the sequence of signs consists ofn − 1 plus signs followed by a minus sign and repeats with periodn.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let fn(x) = r
2 + q
2 +p
2 + · · · +√
2 − x. For x0∈ [0, 2], we define the recurrence xk = fn(xk−1) for k ≥ 1.
We will show that
k→+∞lim xk = 2 cos
π
2n+ 1
.
Let αk = arccos(xk/2) ∈ [0, π/2] then
2 cos(αk) = xk = fn(xk−1) = s
2 + r
2 + q
2 + · · · +p2 − 2 cos(αk−1) is equivalent to
2 cos(2αk) = 4 cos2(αk) − 2 = r
2 + q
2 + · · · +p2 − 2 cos(αk−1) and after n − 1 steps we get
2 cos(2n−1αk) =p2 − 2 cos(αk−1) = 2 sin(αk−1/2) = 2 cos(π/2 − αk−1/2) which implies that
αk =π − αk−1 2n . Hence, as k → +∞,
αk = π
2n −αk−1
2n = π 2n − π
22n +αk−2
22n = π
k
X
j=1
(−1)j−1
2jn +(−1)kα0
2kn → π 2n+ 1, and therefore
xk = 2 cos(αk) → 2 cos
π
2n+ 1
.
