# Convergence of sequence with limit equal to one of its terms

Let $(x_n)$ be a convergent monotone sequence. Suppose that there exists $k ā N$ such that $\lim_{nāā} x_n = x_k$. Show that $x_n = x_k$ for all $n ā„ k$.

So if the sequence is convergent it is bounded. Do I use induction here? Or definitions? I am I confused on how to approach this.

Your sequence is monotone. If it reaches its limit in an actual term, then since each term in the limit is less than or equal to (if monotone increasing), or greater than or equal to (monotone decreasing), the limit of the function, you get by squeezing that $x_k\le x_n \le x_k$ for all $n\ge k$. thus $x_n=x_k$
Just show that if $x_n$ is monotone, then $$\lim_{n\to\infty}x_n\ge x_m \quad \forall m$$ From here, you have that $$\forall n\ge k, \quad x_k\le x_n\le \lim_{n\to\infty}x_n=x_k$$
Suppose the sequence is monotonely increasing. Suppose there exists $m \in \mathbb{N} : x_m > x_k$. Then for every $n > m$ $x_n \ge x_m$ because the sequence is increasing. Then for $n > m$ $|x_k - x_n| \ge |x_k - x_m|$. Then $x_k$ isn't the limit of the sequence. The case with a decreasing sequence is analogous.