Question about limit of subsequence of unbounded sequence I am just wondering if my thoughts on the following make any sort of sense, and if they have the potential to be polished into a valid proof later on.
It is in regard to an exercise in Introduction to Real Analysis by Bartle and Sherbert. $4$th edition.
It asks;

If $(x_{n})$ is an unbounded sequence in $\mathbb{R}$, show that there exists a subsequence $x_{n_k}$ such that $$\lim\frac{1}{x_{n_{k}}} = 0.$$

My general thoughts:
I want to apply the theorem and method of proof found previously in the chapter, which states that every sequence has a monotone subsequence.
It proves that by considering peaks — that is, point $x_{n}$ such that $x_{n} \ge x_{m}$ for all $m \gt n$ — and breaks it into two cases:

*

*infinitely many peaks
In this case we can form a decreasing subsequence consisting of the peaks.


*finitely many peaks
In this case, we consider the index N to be that of the last peak, and then just have $x_{N+1} \le x_{N+2}$ ,.. a strictly  increasing subsequence.
So knowing this , I thought well since $x_{n}$ is unbounded let us consider the a strictly increasing subsequence  (which I wonder if I can show must exist)
I am also wondering, would the following by correct to say;
for all $k \in \mathbb{N}$ there will exist a $n \in \mathbb{N}$ such that $|x_{n_{k}}| \ge k$
and we could prove this by saying, if this weren't true then there would be some $k$ such that no $n$ exists, which would contradict that $x$ itself is unbounded, as the subsequence only takes on values that $x$ can anyways? Would that make sense? and if so then we could use that to avoid the case of the subsequence being equal to zero because we wouldn't consider that point.
Then, using that and the increasing subsequence, we could conclude that its reciprocal is a strictly decreasing sequence, and has a infimum of zero and hence by monotone convergence theorem, its limit is $0$.
Does any of this sound okay? Let me know any suggestions or advice, thanks
 A: Proving that every sequence has a monotone subsequence isn't a proof that every unbounded sequence has an unbounded monotone subsequence (which it does, but you don't need your subsequence to be monotonic - see below).
Your second approach "for all $k \in \mathbb{N}$ there will exist a $n \in \mathbb{N}$ such that $|x_{n_{k}}| \ge k$"is good, and follows immediately by contradiction as you suggest: if not true for some $k$  then $(x_n)$ is bounded with $|x_n| \le k$.
To generate an unbounded subsequence you pick $n_1$ such that $|x_{n_{k}}| > 1$ then recursively define $k+1$ from $k$ by $|x_{n_{k+1}}| > k+1$ and $n_{k+1} \gt n_k$. The second part of the condition is necessary so that your subsequence has the same order of terms as the original sequence. That such an $x_{n_{k+1}}$ exists follows from the same boundedness argument applied to the tail of the sequence: if not, then the tail is bounded and there are a finite number of (finite) terms up to $n_k$ therefore the sequence $(x_n)$ would be bounded.
Now you have a (not necessarily monotonic) unbounded subsequence with all terms $\ne 0$ so you can form the sequence of inverse terms and clearly for $n_k$ with $k> M$ then all $1/|x_{n_{k}}| \lt 1/M$. So the sequence converges to $0$.
A: You ask if it is correct to say that $\forall k>0 \exists n_k$$ (|x_{n_k}|>k)$. Yes. This is just the statement that $(x_n)_{n\in N}$ is an unbounded sequence. Equivalently, $(z_n)_{n\in N}$ is bounded iff $\exists k \forall n (|x_n|<k)$. First let $S^+=\{n :x_n>0\}$ and $S^-=\{n: x_n<0\}$. There cannot exist a positive pair $k_+,k_-$  such that $\forall n\in S^+ (|x_n|<k_+)$ and $\forall n\in S^- (|x_n|<k_-)$, as this would imply $\forall n (|x_n|<k=\max (k^+,k^-) $, which  would make the series $(x_n)_{n\in N}$ bounded. So choose $S\in \{S_+,S_-\}$ such that  $\forall k>0 \exists n\in S (|x_n|>k$. Note that $S$ cannot be a finite set. Now  put $k_1=1$ and let $n_1$ be the least $n\in S$ such that $|x_n|>k_1=1$.  Inductively let $k_{j+1}=1+\max \{|x_m| : m\leq n_j\}$ and let $n_{j+1}$ be the least $n\in S$ such that $|x_n|>k_{j+1}$. Observe that  $n\leq n_j \implies |x_n|<k_{j+1}<|x_{j+1}|\implies x_n\ne x_{j+1} $, so $n_{j+1}>n_j.$ We have $|x_{n_{j+1}}|>|x|_{n_{j}}$ , and  inductively we have $|x_{n_j}|>j$.  The terms of the subsequence $(x_{n_j})_{j\in N}$ are all positive or all negative so it is strictly monotonic ,and $\lim_{j\to \infty}1/x_{n_j}=0$. 
