Answer to Question #123121 in Real Analysis for Wachira Ann Wangari

Question #123121
Show that the sequence below is monotone,bounded and hence find the limit xn=n+1/n
1
Expert's answer
2020-06-22T13:09:58-0400

"x_n=\\frac{n+1}{n}, \\{ 2, 1\\frac{1}{2}, 1\\frac{1}{3}, 1\\frac{1}{4}, 1\\frac{1}{5}, ...\\}."

(i) To prove that the sequence "x_n" is monotonic, let's show that it is always decreasing.

"x_n=\\frac{n+1}{n}, x_{n+1}=\\frac{(n+1)+1}{n+1}=\\frac{n+2}{n+1},"

"x_{n+1}-x_n=\\frac{n+2}{n+1}-\\frac{n+1}{n}=\\frac{n(n+2)-(n+1)^2}{n(n+1)}="

"=\\frac{n^2+2n-n^2-2n-1}{n(n+1)}=-\\frac{1}{n(n+1)}<0."

Hence "x_{n+1}<x_n" "\\forall n" and the sequence "x_n" is decreasing and monotonic.

(ii) A sequence is bounded if it's bounded above and below.

As the sequence is decreasing, then "\\forall n" "x_n \\le x_1=2". 2 is the upper bound.

"x_n=\\frac{n+1}{n}=1+\\frac{1}{n}>1" "\\forall n". 1 is the lower bound.

For all n: "1<x_n\\le2" , so the sequence "x_n" is bounded.

(iii) The sequence "x_n" is monotonic and bounded, so it is convergent and we can find the limit:

"\\lim_{n\\to\\infty} x_n=\\lim_{n\\to\\infty} \\frac{n+1}{n}=\\lim_{n\\to\\infty} (1+\\frac{1}{n})="

"=1+\\lim_{n\\to\\infty} (\\frac{1}{n})=1+0=1."

"\\lim_{n\\to\\infty} x_n=1."


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