Solution to Problem 269

Correct solutions came from Jean Moreau de Saint-Martin, France; Farid Lian, Colombia; Aaron Kahn, Bradley University; Jerome Cherry, Canada; Ahron Teitelman, Israel; John Snyder, USA; Abdul Huq, Oman; Jens Voss, Germany; Philippe Fondanaiche, France; Claudio Baiocchi, Italy; Cee Ann Franklin, USA; Vincent Lynch, UK; Stefan Gatachiu, Romania; Ron Welch, USA; Bill Webb, USA; Dan Dima, Romania; Kai Grinde Myrann, Norway; Lou Cairoli, USA; Brian Bradie, USA; Evan Hagenaars, Bradley University; Allen Druze, USA.

The sequence does, in fact, converge. Its limit is the square root of 2.

One way to show this is with an application of the theory of recurrence relations. Combining the two recurrences, we get d_k+1=2d_k+d_k-1. This is linear and homogeneous, so its solutions are linear combinations of exponential functions. If we try d_k=r^k, we find r²-2r-1=0, which we can solve using the quadratic formula. Call the larger root R and the smaller root S. Then d_k=aR^k+bS^k for some a and b. Notice that n_k/d_k=1+d_k-1/d_k. Since |R|>1 and |S|<1, as k goes to infinity, this goes to 1+1/R, which is the square root of 2.

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ã2007 Alberto L. Delgado