Problem 11890
(American Mathematical Monthly, Vol.123, February 2016) Proposed by G. Apostolopoulos (Greece).
Find allx ∈ (1, +∞) such that
∞
X
k=0
1 2k + 1
1
x2k+1 + x − 1 x + 1
2k+1!
= 1 2
Z x
0
√ dt 1 + t2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
It si known that
∞
X
k=0
z2k+1 2k + 1 =1
2ln 1 + z 1 − z
for |z| < 1, Z x
0
√ dt
1 + t2 = ln(x +p
1 + x2), for x ∈ R.
Hence if x > 1 then 0 < 1/x < 1, 0 <x−1x+1 < 1 and the equation is equivalent to 1
2ln 1 + 1/x 1 − 1/x
+1
2ln 1 + (x − 1)/(x + 1) 1 − (x − 1)/(x + 1)
= 1
2ln(x +p 1 + x2)
or x + 1
x − 1· x = x +p
1 + x2, x4− 2x3− 2x2− 2x + 1 = 0 By letting x = t2+ t with t > 1/ϕ and ϕ = (1 +√
5)/2 (the golden ratio), we have that x4− 2x3− 2x2− 2x + 1 = (t4+ 4t3+ 5t2+ 2t − 1)(t4− t2− 1).
Now P (t) := t4+ 4t3+ 5t2+ 2t − 1 is a strictly increasing function for t > 0, therefore P (t) > P (1/ϕ) > 2
ϕ− 1 > 0, for t > 1/ϕ.
On the other hand t4− t2− 1 = 0 for t > 1/ϕ if and only if t =√ϕ which implies that the required equation has a unique solution in (1, +∞), namely
x = t2+ t = ϕ +√
ϕ ≈ 2.890053638.
