pan which implies that 1 1 − an+1 = 1 1 − an + p = 1 1 − x+ pn

Problem 11559

(American Mathematical Monthly, Vol.118, March 2011) Proposed by Michel Bataille (France).

For positive p and x ∈ (0, 1), define the sequence {xn}n≥0by x0= 1, x1= x, and, for n ≥ 1,

xn+1= pxn−1xn+ (1 − p)x²_n (1 + p)xn−1− pxn

.

Find positive real numbers α, β such that lim_n→∞n^αx_n= β.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will prove that

α = 1

p and β =

Γ

1 p(1−x)

 Γ

x p(1−x)

 . By letting an+1= xn+1/xn, it is easy to see that

a_n+1= p + (1 − p)an

(1 + p) − pa_n which implies that

1

1 − a_n+1 = 1

1 − a_n + p = 1

1 − x+ pn.

Hence

xn =

n−1

Y

k=0

 1 − z

c + k



where z = 1/p > 0 and c = z/(1 − x) > 0. We note that

n→∞lim n^z

n−1

Y

k=0

e^−c+kz = lim

n→∞e^z(^{log n−Pn−1}_{k=0 c+k}¹ ) = e^zΨ(c)

where Ψ is the Digamma function (remember that Ψ(x+1)−Ψ(x) = 1/x and Ψ(x) = log x+O(1/x)).

Finally, by using the Mellin’s Formula, we obtain

n→∞lim n^αxn= lim

n→∞n^zxn= e^zΨ(c) lim

n→∞

n−1

Y

k=0

 1 − z

c + k



e^c+kz = Γ(c) Γ(c − z) = β.