Thus, pn= Fn−1and qn = Fn, where Fn is the nth Fibonacci number defined by the recurrence F0= 0, F1= 1, and Fn+1= Fn+ Fn−1for n &gt

Problem 11716

(American Mathematical Monthly, Vol.120, June-July 2013) Proposed by Oliver Knill (USA).

Let α = (√

5 − 1)/2. Let pn and qnbe the numerator and denominator of the nth continued fraction convergent to α. Thus, pn= Fn−1and qn = Fn, where Fn is the nth Fibonacci number defined by the recurrence F0= 0, F1= 1, and Fn+1= Fn+ Fn−1for n > 1. Show that

√ 5

 α −pn

q_n



=

∞

X

k=0

(−1)^(n+1)(k+1)Ck

qn^2k+25^k , where Ck denotes the kth Catalan number, given by Ck= _k+1^{1 2k}_k
.

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

Since the generating function for the Catalan numbers is

f (z) :=

∞

X

k=0

C_kz^k =1 −√ 1 − 4z

2z for |z| < 1/4, it follows that

∞

X

k=0

(−1)^(n+1)(k+1)Ck

q^2k+2n 5^k = (−1)ⁿ⁺¹

q_n² f (−1)ⁿ⁺¹ q²_n5



=5q_n−p25q²_n+ 20(−1)ⁿ 2qn

because |(−1)ⁿ⁺¹/(q_n²5)| ≤ 1/5. Hence, we have to verify that

√ 5

(√

5 − 1)qn− 2pn



= 5qn−p

25q_n²+ 20(−1)ⁿ, that is

qn+ 2pn=p

5q_n²+ 4(−1)ⁿ.

Now by letting p_n = F_n−1 and q_n = F_n, the above equation is equivalent to F_n−1² − F_n²+ F_n−1F_n= F_n−1F_n− F_n−2F_n+1= (−1)ⁿ,

which is a well known identity.