# 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.

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Well, take any decreasing sequence $a_n$ which converges to $a$ and define $b_{2n}= a_n$ and $b_{2n+1} = a_n + 1/n$. – Rijul Saini Aug 13 '12 at 14:55
A constant sequence will do it. – enzotib Aug 13 '12 at 14:56
In general, if two sequences $\{a_i\}$ and $\{b_i\}$ both converge to $a$, then so does $\{a_1,b_1,a_2,b_2,...\}$. But one sequence might converge "slower" than the other. – Thomas Andrews Aug 13 '12 at 15:00
Thank you everyone, but especially @enzotib – Moderat Aug 13 '12 at 15:03
Other than my obvious comment/answer, of all user suggesting particular sequence, the most interesting (and general) comment is due to @ThomasAndrews. – enzotib Aug 13 '12 at 15:06

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 $0$.

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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 $$a_n=\frac{1}{2^{n+(-1)^n}}.$$

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Or even simpler: $1,0,\frac12,0,\frac14,0,\frac18,0, \dotsc$ – kahen Aug 13 '12 at 15:00

If I'm not wrong your sequence $(*)$ is simply $$a_n = \begin{cases} 2^{3-k} & \text{ if } n = 2k+1 \\ 2^{-k} & \text{ if } n=2k \end{cases}$$

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Oh ok I don't know why it didn't occur to me that it is possible to define terms of $\{a_n\}$ this way. Thanks! – Moderat Aug 13 '12 at 15:02
You are welcome. – Romeo Aug 13 '12 at 15:03

How about $a_n = a+e^{-n}$ if $n$ is even and $a_n = a + \frac{1}{n}$ otherwise?

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