If $a_n$ > 0 and $(a_n)$ from n=1 to ∞ is not bounded above, then there exists an increasing subsequence $(b_n)$ such that $1/b_n$ converges to zero.
My attempt:
Let $(a_n)$ be an increasing sequence not bounded above. Since $(a_n)$ has no upper bound then $sup(a_n)$ =+∞. Since the sequence starts at 1, we know that a lower bound exists.
By the Balzano Weierstrass theorem, since a lower bound exists, we know that there exists a sub-sequence that converges. If we let $(b_n)$ be such sequence, Our goal is to prove that $1/b_n$ converges to zero.
I don't know where to go from there
Best Answer
Unfortunately, this attempt is not right. There is no reason to assume that $a_n$ is increasing. Your application of Bolzano-Weierstrass is incorrect, since that theorem requires $\{a_n\}$ to be bounded, not just bounded below. Finally, if $b_n$ is convergent (to a real number), then $1/b_n$ cannot converge to zero. It converges to $1/b$, where $b = \lim_n b_n$. That's not zero.
What you need is a sequence that runs off to infinity monotonically. One way to construct this is to take the $b_1$ as the first term of $\{a_n\}$ that exceeds $1$. Then let $b_2$ be the next term in the original sequence that exceeds both $2$ and $b_1$. Then let $b_3$ be the next term in the original sequence that exceeds both $3$ and $b_2$.....
Convince yourself that this process is valid.