36th VMO 1998

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Problem A1

Define the sequence x1, x2, x3, ... by x1 = a ≥ 1, xn+1 = 1 + ln(xn(xn2+3)/(1 + 3xn2) ). Show that the sequence converges and find the limit.

 

Solution

(x-1)3 ≥ 0, so x(x2+3)/(1+3x2) ≥ 1. So the ln term is well-defined and non-negative and 1 + ln(x(x2+3)/(1+3x2)) ≥ 1. So by a trivial induction all xn ≥ 1.

Also x2 ≥ 1 implies (x2+3)/(1+3x2) ≤ 1, so x(x2+3)/(1+3x2) ≤ x and hence 1 + ln(x(x2+3)/(1+3x2)) ≤ 1 + ln x ≤ x (*). So the sequence is monotonically decreasing and bounded below by 1. So it must converge. Suppose the limit is L. Then L = 1 + ln(L(L2+3)/(1+Lx2)). But we only have equality in (*) at 1. Hence L = 1.

 


 

36th VMO 1998

© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected/updated 7 Mar 04