Find a sequence $\{a_n\}$ that converges to $a$ without getting "closer and closer" So I have recently come to understand the mechanics behind statements like "Prove that $a_n > b$ for all but finitely many $n$," but I am struggling with this question. 

Find a sequence $\{a_n\}$ and $a \in \mathbb R$ so that $a_n \to a$ but so that the inequality $$|a_{n+1} - a| < |a_n-a|$$ is violated for infinitely many $n$. 

The moral we are supposed to draw from a question like this is  that the sequence can converge without getting "closer and closer" but that is not what I draw from this statement. 
I see this question as talking about a sequence like one from this pdf which looks like this 

$8, 1, 4, 1/2, 2, 1/4, 1, 1/8, 1/2, 1/16, \ldots \tag{*}$

This sequence "bounces around" like I would imagine the one we are supposed to come up with for the question, but I don't know what the explicit formula that generates the one in $(*)$. 
So I'm a little stuck. Any starting hints would really help.
 A: Consider the sequence
$$\frac{1}{2}, 1, \frac{1}{8},\frac{1}{4}, \frac{1}{32}, \frac{1}{16}, \frac{1}{128}, \frac{1}{64},\dots.$$ 
This converges to $a=0$. 
A precise description of this sequence $a_0, a_1, a_2,\dots$ could be $a_{n}=\frac{1}{2^{n+1}}$  if $n$ is even and $a_n=\frac{1}{2^{n-1}}$ if $n$ is odd.
If we want a really explicit formula, let 
$$a_n=\frac{1}{2^{n+(-1)^n}}.$$
A: When trying to construct an example, it’s usually a good idea to keep things as simple as possible, so let’s try for one with $a=0$. Now you want a sequence $\langle a_n:n\in\Bbb N\rangle$ that converges to $0$ but is such that there are infinitely many $n$ with $a_{n+1}$ further away from $0$ than $a_n$ is. If, still in the interests of keeping things simple, we try to make all of the terms $a_n$ positive, this means that we want to have infinitely many $n$ such that $a_{n+1}>a_n$.
Let’s start by setting $a_0=1$. We want the sequence to have limit $0$, so we’ll choose $a_1$ to be closer to $0$; $a_1=\frac12$ seems like a reasonable choice. Now let’s throw in a step backwards, away from $0$; stepping half the way back to our starting point still leaves us with some net progress towards $0$, so let’s try $a_2=\frac34$. Now we want to move towards $0$ again. Our previous forward progress was to $\frac12$; let’s get twice as close by taking $a_3=\frac14$. So far we have 
$$\left\langle 1,\frac12,\frac34,\frac14\right\rangle\;.$$
Now let’s take a second step back, say halfway to our previous best forward progress of $\frac12$: that puts us at $a_4=\frac38$. Then we improve on our previous best of $\frac14$ by setting $a_5=\frac18$, step halfway back to the previous best of $\frac14$ by setting $a_6=\frac3{16}$ and so on. At this point we have
$$\left\langle 1,\frac12,\frac34,\frac14,\frac38,\frac18,\frac3{16},\dots\right\rangle\;.$$
I’ll leave it to you to write down a general formula for $a_n$; you’ll want to split it into two cases, one for even $n$ and one for odd $n$. Notice that whenever $n$ is odd, $a_{n+1}>a_n$, so there are infinitely many $n$ at which the sequence actually steps away from $0$. Still, it’s not hard to see that it does converge to $0$.
A: If I'm not wrong your sequence $(*)$ is simply 
$$
a_n = \begin{cases}
2^{3-k} & \text{ if } n = 2k+1 \\
2^{-k} & \text{ if } n=2k
\end{cases}
$$
A: How about $a_n = a+e^{-n}$ if $n$ is even and $a_n = a + \frac{1}{n}$ otherwise?
