Therefore 1 − k αnk+ αk = αnk+1+ α2k− kαk αnk+1+ α2k = αnk+1+ 1 αnk+1+ α2k =1 + α−k(n+1) 1 + α−(n−1)k

Problem 12110

(American Mathematical Monthly, Vol.126, April 2019) Proposed by P. J. Rodriguez de Rivera and A. Plaza (Spain).

Letα^k = (k +√

k²+ 4)/2. Evaluate

k→∞lim

∞

Y

n=1



1 − k

αⁿk + α^k

 .

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

Solution. We have that

α²k =k²+ k²+ 4 + 2k√ k²+ 4

4 = kα^k+ 1.

Therefore

1 − k

αⁿk+ α^k = αⁿk⁺¹+ α²k− kα^k αⁿk⁺¹+ α²k

= αⁿk⁺¹+ 1 αⁿk⁺¹+ α²k

=1 + α⁻k⁽ⁿ⁺¹⁾

1 + α⁻⁽ⁿ⁻¹⁾k

.

Finally, since α^k> 1 for k > 0, and lim^k→∞α⁻k¹= 0, we may conclude that

k→∞lim

∞

Y

n=1



1 − k

αⁿk + α^k



= lim

k→∞ lim

N →∞

N

Y

n=1

1 + α⁻⁽ⁿ⁺¹⁾k

1 + α⁻k⁽ⁿ⁻¹⁾

= lim

k→∞ lim

N →∞

(1 + α^−Nk )(1 + α^{−(N +1)}k ) (1 + α⁰k)(1 + α⁻¹k )

= lim

k→∞

1

2(1 + α⁻¹k )= 1 2.