# If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent. [duplicate]

If $$(x_n)$$ is monotone and contains a convergent subsequence $$(x_{n_i}),$$ then $$(x_n)$$ is convergent.

Demonstrating an alternative proof, please provide feedback :) Thank you.

Suppose $$(x_n)$$ is monotone and contains a convergent subsequence $$(x_{n_i}).$$

Given that $$(x_{n_i})$$ is convergent, it is bounded above by some upper bound $$b \in \mathbb{N},$$ that is, $$x_{n_i} \leq b,$$ $$\forall i \in \mathbb{N}.$$

Suppose $$(x_n)$$ is divergent. Given its monotonicity, it follows that $$(x_n)$$ is unbounded, that is, $$\forall M \in \mathbb{N},$$ $$\exists N \in \mathbb{N}$$ such that $$\forall n \geq N$$ it follows that $$x_n > M$$ (at some point "$$N$$," the sequence passes the boundary $$M$$ for any boundary $$M \in \mathbb{N}).$$ If $$(x_n)$$ is bounded, it is necessarily the case that $$(x_n),$$ given that it is monotone increasing, IS convergent (you can prove this).

So, $$\exists N \in \mathbb{N}$$ such that $$\forall n \geq N$$ it follows that $$x_n > b.$$

Given that $$(x_{n_i})$$ is bounded above by $$b,$$ this means that $$\forall i \in \mathbb{N},$$ $$n_i < N.$$

Hence, as by definition of a subsequence it is the case that $$n_1 < n_2 < \cdots n_i,$$ the subsequence $$(x_{n_i})$$ contains fewer than $$N$$ elements (at most $$N - 1$$ elements).

However, a subsequence is defined as a function whose domain is the natural numbers. Given that $$(x_{n_i})$$ contains fewer than $$N$$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.

Therefore, $$(x_n)$$ is convergent.

• Misfire in my brain. Fixed it. Jan 28 '19 at 23:14
• Thanks for pointing that out... lol Jan 28 '19 at 23:14
• I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer. Jan 28 '19 at 23:14
• Ok. I'll change it. Jan 28 '19 at 23:15

If $$x_n$$ did not converge then, by monotonicity, $$x_n$$ would be unbounded, in particular $$x_n - 1 > x = \lim x_{n_i}$$ for all $$n \geq n_0,$$ for some $$n_0.$$ But there exists a firs $$i_0$$ such that $$n_i \geq n_0$$ for all $$i \geq i_0,$$ and so $$x_{n_i} \geq x + 1$$ for all $$i \geq i_0,$$ a contradiction. Q.E.D.