Subsequences and Convergence 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? 
 A: For the first question, since every sequence contains either an increasing or a decreasing subsequence, every bounded sequence contains a convergent subsequence. So if you've shown it is bounded, you are done.
For the second question, I don't think you are correct in saying that $(a_n-a)$ is a null sequence; and you never said who $a_{n_k}$ was, so the paragraph is empty.. Now you are saying $a_{n_k}$ is an arbitrary subsequence; well, it doesn't follow then. Take $a_n = \frac{1}{n}$ if $n$ is even, $a_n = n$ if $n$ is odd, and take $a=0$. Then for every $\epsilon\gt 0$ we have that $|a_n - 0|\lt \epsilon$ frequently, but $(a_n-0)$ is not a null sequence, and there are plenty of subsequence $(a_{n_k})$ for which $(a_{n_k}-0)$ is not a null sequence either. For example, $n_k=2k+1$ gives the sequence $(2k+1)_{k=1}^{\infty}$, which is certainly not a null sequence. 
Instead, for (a)$\Rightarrow $(b), construct the subsequence inductively; pick $n_1$ so that $|a_{n_1}-a|\lt 1$; then pick $n_2\gt n_1$ such that $|a_{n_2}-a|\lt \frac{1}{2}$; then pick $n_3\gt n_2$ such that $|a_{n_3}-a|\lt \frac{1}{4}$; etc. Of course, the condition you have in (a) is what you need to justify this is possible. 
You never did anything about (b)$\Rightarrow$(a), however. If you have a subsequence $(a_{n_k})$ that converges to $a$, you need to prove that for every $\epsilon\gt 0$, $|a_n-a|\lt\epsilon$ frequently; it's easy, but you haven't done it.
Your attempted conclusion at the end does not follow; what makes you think that if $|a_n-a|\lt \epsilon$ from some point forward, then $|a_n-a|\geq \epsilon$ "frequently"? To occur frequently, it must first of all occur infinitely many times, but here it can only occur $N-1$ times at the most.
Think about the condition you wrote for the "ultimately" clause. It should look very familiar!
