2
$\begingroup$

I encountered a problem in my book that I'm not sure how solve. The problem is as follows:

{$x_n$} is a monotone decreasing sequence of real numbers. Suppose there exists a k in N such that lim$_{n\to\infty}$ $x_n$ = $x_k$. Prove that $x_n$ = $x_k$ for all n > k.

What would be the general approach to solving a problem like this?

  • 0
    I would assume that there exists a number $n>k$ with $x_n – 2017-02-23

2 Answers 2

0

Applying the definition of limit you find

$\forall\varepsilon>0,\exists m\in\mathbb{N}$ such that $\forall n>m$, $|x_n-x_k|<\varepsilon$.

And hence

$x_k-\varepsilon\leq x_n\leq x_k+\varepsilon$

but since $(x_n)$ is decreasing the only possibility for $x_k-\varepsilon\leq x_n$ to hold is $x_k=x_n$.

0

Since $(x_n)$ is decreasing,

$\lim_{n \to \infty}x_n= \inf \{x_1,x_2,x_3,...\}$.

Hence $x_k=\inf \{x_1,x_2,x_3,...\}$.

For $n>k$ we then have:

$x_k \le x_n \le x_k$

( the reason for the second $\le$: $(x_n)$ is decreasing).