By the recurrence relation xn+1= Pn−1 k=0sin(xk

Problem 11995

(American Mathematical Monthly, Vol.124, August-September 2017) Proposed by D. S. Marinescu and M. Monea (Romania).

Suppose0 < x0< π, and for n ≥ 1 define xⁿ= n¹

Pⁿ⁻1

k=0sin(x^k). Find lim

n→∞

xnpln(n).

Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.

Solution. By the recurrence relation

x_n+1= Pⁿ⁻1

k=0sin(x^k) + sin(xⁿ)

n+ 1 =nxn+ sin(xⁿ) n+ 1 which implies that for 0 < x0< π,

0 < xⁿ+1< nxn+ xⁿ n+ 1 = xⁿ

that is the sequence (xⁿ)ⁿ is positive and decreasing and therefore it has a limit l which satisfies (n + 1)l = nl + sin(l). Thus it follows that l = 0. Moreover, by Taylor approximation,

xn+1= nxn+ xⁿ−^x

3 n

6 + o(x³n)

n+ 1 = xⁿ



1 − x²_n

6(n + 1)+o(x²n) n+ 1



and 1 x²_n

+1

= 1 x²_n



1 − x²_n

6(n + 1)+o(x²n) n+ 1

⁻2

= 1 x²_n



1 + x²_n

3(n + 1)+o(x²n) n+ 1



= 1

x²_n + 1

3(n + 1)+ o(1) n+ 1. Finally, by the Stolz-Cesaro Theorem,

n→∞lim(xⁿpln(n))²= lim

n→∞

ln(n)

1 x²

n

SC= lim

n→∞

ln(n + 1) − ln(n)

1 x²

n+1−x¹²

n

= lim

n→∞

ln(1 +n¹)

1

3(n+1)+n^o(1)+1

= 3

Since, xnpln(n) ≥ 0, we may conclude that lim_n→∞x_npln(n) =√

3.