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Let $\{x_n\}$ be a convergent monotone sequence. Suppose that there exists a $k \in N$ such that $$ \lim_{n \rightarrow \infty} x_n = x_k $$ Show that $x_n = x_k$ for all $n \ge k$. I am completely lost in knowing where to begin this problem.

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2 Answers 2

Hint: suppose $x_m\neq x_k$ for some $m>k$. Can your sequence still be monotone?

Solution: Suppose $\displaystyle\lim_{n\rightarrow\infty}x_n=x_k$ for some $k$. Now, suppose that there exists an $m>k$ such that $x_m\neq x_k$ - say $x_m<x_k$, WLOG. Set $\epsilon=x_k-x_m$. Now, because the sequence converges to $x_k$, there must be a $j>m>k$ such that $\vert x_k-x_j\vert<\epsilon$, or written another way, $$x_k-\epsilon<x_j<x_k+\epsilon.$$ By construction, $x_m=x_k-\epsilon$, and so we have $x_j>x_m$. However this means that the sequence cannot be monotone, since for $k<m<j$ we have $x_k>x_m$ while $x_m<x_j$ (i.e. the sequence decreases, then increases again). The same proof will work if we supposed that $x_m>x_k$.

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I still seem to be completely lost in getting anywhere in this problem. If the sequence cannot still be monotone, what does this then mean? –  user72195 Jun 26 '13 at 12:40
    
We want to prove a statement like "if the sequence is convergent and monotone, then X is true". By contrapositive, if "not X" implies the sequence is either not convergent or not monotone, the original statement is proved. –  icurays1 Jun 26 '13 at 21:21
    
I am not understanding how the if "not X" will contradict the sequence being monotone. I have that the sequence is monotone and decreasing so the limit = xk is the greatest lower bound. But how is this contradicted if Xn is not equal to Xk? –  user72195 Jun 27 '13 at 2:08
    
@user72195 I've edited my answer to give a more complete solution. Hope it makes sense to you. –  icurays1 Jun 27 '13 at 18:02

Well, the alternative is that $x_n > x_k$ for some $n > k$. What does this imply with regards to the relation between the limit and $x_k$, given that the sequence is monotone?

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