I came across the following problems on subsequences during the course of my study of real analysis:
True or false: If $(a_n)$ is any sequence in $\mathbb{R}$, then the sequence $x_n = \frac{a_n}{1+|a_n|}$ has a convergent subseqeunce.
We know that $|x_n| < 1$ for all $n$. Hence by the Weierstrass-Bolzano Theorem $(x_n)$ has a convergent subsequence.
If $(a_n)$ is a sequence in $\mathbb{R}$ and $a \in \mathbb{R}$, the following conditions are equivalent. (a) $(\forall \epsilon >0) |a_n-a| < \epsilon$ frequently, (b) there exists a subsequence of $(a_n)$ converging to $a$. What happens if we change "frequently" to "ultimately."? How would (b) change?
So we know that $(\forall \epsilon >0)(\forall N) \ \exists n \geq N \ni |a_n-a| < \epsilon$. In other words, $(a_n-a)$ is a null sequence. Thus from a previous theorem $(a_{n_{k}}-a)$ is a null sequence where $(a_{n_{k}})$ is some arbitrary subsequence.
If we changed the wording to "ultimately" then we would have: $(\forall \epsilon >0) \ \exists N \ni n\geq N \Rightarrow |a_n-a| < \epsilon$. Then $(a_n)$ wouldn't have a subsequence converging to $a$ because $|a_n-a| \geq \epsilon$ frequently?