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

Question

Show that the sequence below is monotone,bounded and hence find the limit xn=n+1/n

Answer 1

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

(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:

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

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

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