# Hard inequalities question

by nokia8650
Tags: inequalities
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,294 If (f) is the part you are having trouble with, then presumably you have already proved that $2x_n^2- (2n-1)x- (n+1)= 0$ (part (e)). Now you want to find the smallest n such that $x_n< n+ 0.05$. You could, for example, solve that using the quadrative formula and compare the solutions to n+ 0.05. Have you calculated some values of $x_n$? What are $x_0$ $x_1$, etc.?