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45 < x_{1000} < 45.1 if x_0 = 5. x_{n+1} = x_n +1/x_n

Source: 1975 Hungary - Kürschák Competition p3

October 15, 2022
recurrence relationSequencealgebrainequalities

Problem Statement

Let x0=5  ,   xn+1=xn+1xn.x_0 = 5\,\, ,\, \,\,x_{n+1} = x_n +\frac{1}{x_n}. Prove that 45<x1000<45.145 < x_{1000} < 45.1.
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