Show that the sequence (fn) sequence where
fn(x)= x/(1+nx^2), x∈[2,∞] is uniformly convergent in [2,∞]
To prove that the given sequence of functions converges uniformly on "\\displaystyle\n[2,\\infty)", it suffices to show that the sequence of functions converges uniformly on "\\displaystyle \\R".
Now,
"\\displaystyle\n\\left|f_n(x)\\right|=\\left|\\frac{x}{1+nx^2}\\right|=\\left|\\frac{x\\sqrt{n}}{(1+nx^2)\\sqrt{n}}\\right|=\\frac{1}{\\sqrt{n}}\\left(\\frac{\\sqrt{n}\\ |x|}{1+{(\\sqrt{n}|x|)}^2}\\right)=\\frac{1}{\\sqrt{n}}\\left(\\frac{t}{1+t^2}\\right)"
where "\\displaystyle\nt=\\sqrt{n}\\ |x|.\\text{ But, }\\frac{t}{1+t^2}\\leq\\frac{1}{2}\\ \\ \\ \\forall\\ t\\in\\R\\ \\text{since } (1\u2212t)^2 \u2265 0,\\ \\Rightarrow 2t \u2264 1 + t^2."
Thus, from the above inequality, we have that; "\\displaystyle\n|f_n(x)|=\\frac{1}{\\sqrt{n}}\\left(\\frac{t}{1+t^2}\\right)\u2264\\frac{1}{\\sqrt{n}}\\left(\\frac{1}{2}\\right)\\rightarrow0,\\ \\text{as } n\\rightarrow \\infty\\ \\ \\ \\forall\\ \\ \\ x\\in\\R.\\\\\n\\text{Hence, given }\\epsilon>0\\ \\text{choose } N = \\frac{1}{4\\epsilon^2}.\\ \\text{Then } |f_n(x)| < \\epsilon\\ \\ \\ \\forall\\ \\ x\\in\\R\\ \\text{if}\\ n>N." Showing that the sequence of functions converges uniformly to "\\displaystyle\n0 \\text{ on }\\R." And since "\\displaystyle\n[2,\\infty)\\subset\\R," then the given sequence of functions converges uniformly to "\\displaystyle\n0 \\text{ on }[2,\\infty)."
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