Let $\left\{{a_n} \right\}$ be an unbound sequence. Prove that thereexists a subsequence $\left\{{a_{p_n}} \right\}$ of $\left\{{a_n}\right\}$ such that $\left\{ \frac{1}{a_{p_n}} \right\}$ converges to$0$.
So this is a problem from an elementary real analysis textbook through which I'm currently working my way up, on my own.
Now, this seems like it should be an easy problem. However, as non-math major from an engineering background working out the details, setting up the sequence, proof write-up, and the like, is a bit of challenge for me. I cannot, for example, be sure whether I have done it correctly (is it generally OK-ish?), or perhaps I should have started off (how?) with some recursive definition.
So, what do you guys think?
Proof. If we prove there exists a subsequence $\left\{a_{p_n} \right\}$ of $\left\{a_{n} \right\}$ with nonzero terms such that itconverges to $ \infty$, we are done. Let $S = \left\{n : a_n =\text{sup} \left\{ a_1, a_2, ..., a_n\right\}, a_n \neq 0\right\} $.The set $S$ is infinite; it can be expressed as a strictly increasingsequence $\left\{p_n \right\}$ which is then used to form thesubsequence $\left\{a_{p_n} \right\}$. Now, we prove $\left\{a_{p_n}\right\}$ is an unbound increasing sequence.
To show that it is increasing, we note that, since $\left\{ a_1, a_2,..., a_{p_n}\right\} \subseteq \left\{ a_1, a_2, ...,a_{p_{n+1}}\right\}$, it follows from the definition of $S$ that$a_{p_n} \leq a_{p_{n+1}}$.
To show that it is unbounded, first we prove that $a_{p_n} \geq a_n$.Let n be arbitrary. Since $p_n \in S$, we have $a_{p_n} =\text{sup}\left\{a_1,...,a_{p_n}\right\}$. But $n \leq p_n$, so $a_n\in \left\{a_1,...,a_{p_n}\right\}$. Thus, $a_{p_n} \geq a_n$. Now,let $\left\{a_{p_n} \right\}$ be bounded. Let $M$ be an upper boundfor this. Because $\left\{a_{n} \right\}$ is unbounded, for some $N\in \mathbb{Z^+}$, we have $a_N > M$. Since $a_{p_N} \geq a_N$, itfollows that $a_{p_N} > M$. Therefore, $\left\{a_{p_n} \right\}$,too, is unbounded.