# $(a_n)$ is a sequence s.t. its partial limits are $3 , - 1$. Define $b_n = | a_n - 1 |$. Show $b_n \rightarrow 2$

Problem: Let $$(a_n)$$ be a sequence s.t. its partial limits are only from the set $$\{ 3 , - 1 \}$$ ( infinite limits are not allowed to be partial limits of $$(a_n)$$ ). Define a new sequence $$b_n = | a_n - 1 |$$. Show that $$b_n \rightarrow 2$$.

Attempt: Suppose that $$b_n \nrightarrow 2$$. Meaning there exists $$\epsilon>0$$ s.t. $$\forall N \in \mathbb{N} . \exists n^*>N.| |a_n - 1 | - 2 | \geq \epsilon$$. Choose $$N=1$$, then there exists $$n^* > N$$ s.t. $$| |a_n^* - 1 | - 2 | \geq \epsilon \iff | |a_n^* - 1 | - 2 | \geq \epsilon \iff |a_n^* - 1 | - 2 \geq \epsilon$$ or $$|a_n^* - 1 | - 2 \leq - \epsilon$$.
[ I am stuck. I don't see how I can continue if I assume wlog either $$|a_n^* - 1 | - 2 \geq \epsilon$$ or $$|a_n^* - 1 | - 2 \leq - \epsilon$$, I think a better option would be If I wouldn't try to prove by contradiction, but that attempt was failure for me as well ].

Do you have any ideas how I should continue? or if I don't choose to prove by contradiction, how would one proceed after taking arbitrary $$\epsilon > 0$$ and using the givens that $$3 ,-1$$ are partial limits?

Edit: The teacher-assistant told me that the theorem is true.

• Are $\pm\infty$ allowed as "partial limits"? If not this is false (consider the sequence $3, -1, 1, 3, -1, 2, 3, -1, 3\dots$.) Apr 25, 2021 at 11:34
• You are right, I edited the question so that the only partial limits are 3 and -1. If so then the sequence $(a_n)$ is bounded. But how does this help me? Apr 25, 2021 at 12:20
• Adding the word only does not clarify whether you're allowing infinite "partial limits". Apr 25, 2021 at 12:22
• @Snoop Exactly. Apr 25, 2021 at 12:49
• @Snoop Yes, and those two limit points are : $3 , -1$. That is how the original question appears but only after looking at the answer I could solve the ambiguities of the question. Apr 25, 2021 at 14:35

Note that since there are no infinite partial, or sub-sequential limits, then $$\left(a_n\right)_{n\in\mathbb{N}}$$ is bounded.

We claim that for all $$\varepsilon>0$$, there exists an $$N\in\mathbb{N}$$ such that for all $$n\geq N$$, we have $$|a_n-(-1)|<\varepsilon$$ or $$|a_n-3|<\varepsilon$$.

Suppose, for contradiction, that there exists an $$\varepsilon>0$$ such that for all $$N\in\mathbb{N}$$, there exists an $$n\geq N$$ with $$|a_{n}-(-1)|\geq\varepsilon$$ and $$|a_{n}-3|\geq\varepsilon$$. After possibly permuting or re-ordering terms, this gives rise to a sub-sequence $$\left(a_{n_k}\right)_{k\in\mathbb{N}}$$.

Since this sub-sequence $$\left(a_{n_k}\right)_{k\in\mathbb{N}}$$ is bounded, then by Bolzano-Wierstrass, there is a convergent sub-sequence of it, which is a partial limit of $$\left(a_n\right)_{n\in\mathbb{N}}$$. This new partial limit cannot be $$-1$$ or $$3$$, which is a contradiction, thus the claim holds.

Define the sequence $$b_n=|a_n-1|$$. We want to show that $$\lim\limits_{n\to\infty}b_n=2$$.

Let $$\varepsilon>0$$. Then there exists an $$N\in\mathbb{N}$$ such that for all $$n\geq N$$, we have $$|a_n-(-1)|<\varepsilon$$ or $$|a_n-3|<\varepsilon$$.

Let $$n\geq N$$. Now, there are two cases: $$a_n-1\geq 0$$ or $$a_n-1<0$$. In the former case, we have that $$|b_n-2|=||a_n-1|-2|=|a_n-3|$$ In the latter case, we have that $$|b_n-2|=||a_n-1|-2|=|-a_n-1|=|a_n-(-1)|$$ Now $$|b_n-2|<\varepsilon$$, thus $$\lim\limits_{n\to\infty}b_n=2$$.

Edit 2: To make the cases more clear, picking up from "We want to show that $$\lim\limits_{n\to\infty}b_n=2$$", this is equivalent to showing that for all $$n\geq N$$, we have $$||a_n-1|-2|<\varepsilon\Longleftrightarrow-\varepsilon<|a_n-1|-2<\varepsilon \Longleftrightarrow 2-\varepsilon < |a_n-1|<2+\varepsilon$$

In the case where $$a_n-1\geq 0$$, we have $$3-\varepsilon The case when $$a_n-1<0$$ is similar.

Edit 1: I did not read the comments and I misinterpreted the statement of the question previously which lead to the answer below.

I think this statement is false. Consider the sequence $$\left(a_n\right)_{n\in\mathbb{N}}$$ defined as

$$\begin{cases} n &\text{ if }n\equiv 1\mod 3\\ -1&\text{ if }n\equiv 2\mod 3\\ 3&\text{ if }n\equiv 0\mod 3 \end{cases}$$

The only finite, partial limits of $$\left(a_n\right)_{n\in\mathbb{N}}$$ are $$-1$$ and $$3$$.

The sequence $$b_n=|a_n-1|$$ is seen to be $$b_n = \begin{cases} n-1 &\text{ if } n\equiv 1\mod 3\\ 2 &\text{ if else } \end{cases}$$ but $$\left(b_{n}\right)_{n\in\mathbb{N}}$$ does not converge, since $$\limsup\limits_{n\to\infty}(b_n)=\infty$$ and $$\liminf\limits_{n\to\infty}(b_n)=2$$.

• $a_{3k + 1} \to \infty$ and that's not allowed as the OP clarified. Apr 25, 2021 at 14:52
• I don't understand the line:" there are two cases, $|a_n-1|\geq 0$ or $|a_n-1|<0$ ". Where do we derive those cases from?. Apr 25, 2021 at 16:35
• @hazelnut_116 I just edited. The $|a_n-1|<0$ didn't even make sense. You derive those cases because we want to look at $|b_n-2|=||a_n-1|-2|$. There are two cases: $a_n-1\geq 0$ or $a_n-1<0$ for some fixed $n\geq N$. Apr 25, 2021 at 16:37
• I see, but from where do we derive those cases ? Apr 25, 2021 at 16:38
• @hazelnut_116 after playing around with it a little bit, I noticed that the sequence had to be bounded. Then, if there were anything else that $a_n$ was close to besides $-1$ and $3$ for large $n$, it would have to be a limit point, as the sequence is bounded. Apr 25, 2021 at 17:17