If $(x_{2n})_n, (x_{2n+1})_n$ and $(x_{3n})_n$ converges, prove that $(x_n)_n$ also converges.

Question

let's say $(x_n)_n$ is a row in $\mathbb{R}$. If $(x_{2n})_n, (x_{2n+1})_n$ and $(x_{3n})_n$ converges, I want to prove that $(x_n)_n$ also converges.

I think it's normal because $(x_n)_n$ exist out of the three rows, but I can't prove it. Can someone help me with this?

What do you mean by "a row in R"? I've never seen that notation before. — RobertTheTutor, Apr 10 '21 at 11:56
If the limits are $l,m$ and $p$ show rht $l=m=p$ by considering $(x_{2(3n)})$, $(x_{2(3n+1+1)})$ and $(x_{3(2n)})$. — Kavi Rama Murthy, Apr 10 '21 at 12:05
@RobertTheTutor My guess is that this is a translation from the Dutch language. — J. De Ro, Apr 10 '21 at 12:18
yes i'm sorry it's a dutch translation, i don't know how you say it in englisch otherwise? — questmath, Apr 10 '21 at 12:22
Another one: https://math.stackexchange.com/q/3409684/42969. — Martin R, Apr 10 '21 at 12:58

score 3 · Answer 1 · answered Apr 10 '21 at 12:06

Suppose that $(x_{2n})$ converges to $R$, $(x_{2n+1})$ converges to $S$ and $(x_{3n})$ converges to $T$.

Consider the sequence $(x_{6n})$: it's a subsequence of $(x_{2n})$, so it converges to $R$. But it's also a subsequence of $(x_{3n})$, so it converges to $T$, showing that $R=T$. Similarly, $(x_{6n+3})$ is a subsequence of both $(x_{2n+1})$ and $(x_{3n})$, showing that $S=T$, and hence $R=S$.

Finally, note that the sequences $(x_{2n})$ and $(x_{2n+1})$ partition the sequence $(x_{n})$ into two disjoint subsequences, both converging to the same limit $R$, implying that $(x_{n})$ converges to $R$.

score 2 · Accepted Answer · answered Apr 10 '21 at 12:10

2

Let $l_1 = \lim\limits_{n \to \infty} x_{2n+1}$, $l_2 = \lim\limits_{n \to \infty} x_{2n}$ and $l_3 = \lim\limits_{n \to \infty} x_{3n}$

It is sufficient to prove that $l_1 = l_2$ which immediately follows from

$$l_1 = \lim\limits_{n \to \infty} x_{2(3n + 1)+1} = \lim\limits_{n \to \infty} x_{3(2n+1)} = l_3 = \lim\limits_{n \to \infty} x_{3(2n)}=\lim\limits_{n \to \infty} x_{2(3n)}=l_2$$

answered Apr 10 '21 at 12:10

mathcounterexamples.net

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why does it implies that if l1=2=l3 , then xn converges? – questmath Apr 10 '21 at 12:23
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Let $l=l_1=l_2$. If $\vert x_{2n+1} - l \vert \lt \epsilon$ for $n \gt N_1$ and $\vert x_{2n} - l \vert \lt \epsilon$ for $n \gt N_2$ then $\vert x_n - l\vert \lt \epsilon$ for $n \gt 2 \sup(N_1,N_2)+1$. – mathcounterexamples.net Apr 10 '21 at 12:27

If $(x_{2n})_n, (x_{2n+1})_n$ and $(x_{3n})_n$ converges, prove that $(x_n)_n$ also converges.

2 Answers2