By the pentagonal number theorem, it suffices to prove that an = 1/Qn−1 k=1(1 − xk) for n &gt

Problem 11659

(American Mathematical Monthly, Vol.119, August-September 2012) Proposed by Albert Stadler (Switzerland).

Let x be real with 0 < x < 1, and consider the sequence {an}n≥0 given by a0= 0, a1= 1, and, for n > 1,

an= a²_n−1

xa_n−2+ (1 − x)a_n−1. Show that

n→∞lim 1 a_n =

∞

X

k=−∞

(−1)^kx^k(3k−1)/2.

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

By the pentagonal number theorem, it suffices to prove that a_n = 1/Qn−1

k=1(1 − x^k) for n > 0. It is easy to verify that this equality holds for n = 1 and n = 2. If n > 2 then, by the inductive hypothesis,

a_n= 1

Qn−2

k=1(1 − x^k)2 · 1

x Qn−3

k=1(1−x^k)+Qn−2^1−x k=1(1−x^k)

= 1

Qn−2

k=1(1 − x^k)· 1

x(1 − xⁿ⁻²) + (1 − x)

= 1

Qn−2

k=1(1 − x^k)· 1

1 − xⁿ⁻¹ = 1 Qn−1

k=1(1 − x^k).