# Show that limit of sequence exists with the help of subsequence.

Let $(x_n)^\infty_{n=1}$ be a sequence of real numbers. Suppose that there is a real number $L$ such that

$$L=\lim_{n\to\infty}x_{3n-1}=\lim_{n\to\infty}x_{3n+1}=\lim_{n\to\infty}x_{3n}$$

Show that $\lim_{n\to\infty}x_n$ exists and equals to $L$.

I dont even have a clue how to start.

I know every bounded sequence of real numbers has a convergent subsequence.

so is it if I can show $\lim_{n\to\infty}x_{3n-1}$ , $\lim_{n\to\infty}x_{3n+1}$ , $\lim_{n\to\infty}x_{3n}$ are convergent $\rightarrow$ I can show $x_n$ is bounded $\rightarrow$ $\lim_{n\to\infty}x_n$ exists?????

As I already know $x_{3n-1}$, $x_{3n+1}$ and $x_{3n}$ converge, is that enough to show $x_n$ converges too????

Thank you!!

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What is the (precise) definition of $\lim\limits_{n\to\infty}x_n=L$? (Note that not all bounded sequences converge.) – Jonas Meyer Feb 3 '13 at 0:16
all I know is for $\lim_{n→∞} x_n=L$ then $|x_n-L|<ε$ – Paul Feb 3 '13 at 0:18
That is not a correct definition. Both $n$ and $\varepsilon$ must be quantified; what you wrote is technically meaningless. I recommend looking up the precise definition of what it means for a sequence to converge to a limit $L$. This is a good exercise in using that definition. – Jonas Meyer Feb 3 '13 at 0:20
You should be able to get something out of the fact that the three subsequences together form the entire sequence, in other words every natural number has one of the forms $3n-1$, $3n+1$, or $3n$. – Harald Hanche-Olsen Feb 3 '13 at 0:21
You mean the definition of $(a_n)^∞_{n=1}$ if for every $ε>0$, there is an integer $N=N(ε)>0$ such that $|x_n-L|<ε$ for all $n≥N$ ??? – Paul Feb 3 '13 at 0:27

Your main problem is that you’ve not yet correctly absorbed the definition of limit. The statement that $\lim_{x\to\infty}x_n=L$ does not mean that $|x_n-L|<\epsilon$. This actually isn’t even a meaningful statement. What is $\epsilon$? For what values of $n$?
The statement that $\lim_{x\to\infty}x_n=L$ means that for each $\epsilon>0$ there is some natural number $m_\epsilon$ such that $|x_n-L|<\epsilon$ whenever $n\ge m_\epsilon$. To show that $\lim_{x\to\infty}x_n=L$, you must show that no matter how small a positive number $\epsilon$ I choose, you can find some $m_\epsilon\in\Bbb N$ such that all of the terms $x_{m_\epsilon},x_{m_\epsilon+1},x_{m_\epsilon+2},\dots$ are less between $L-\epsilon$ and $L+\epsilon$. We sometimes say that by taking a tail of the sequence starting far enough out, we can ensure that the whose tail is within $\epsilon$ units of $L$.
In this problem you’re given that the sequence $\langle x_{3n}:n\in\Bbb N\rangle$ converges to $L$, i.e., that $\lim_{x\to\infty}x_{3n}=L$. According to the definition above, that means that for each $\epsilon>0$ there is an $m_\epsilon^{(0)}\in\Bbb N$ such that $|x_{3n}-L|<\epsilon$ whenever $n\ge m_\epsilon^{(0)}$. (The superscript is not an exponent; it’s just a label.)
You’re also given that $\lim_{x\to\infty}x_{3n+1}=L$, which means that for each $\epsilon>0$ there is some $m_\epsilon^{(1)}\in\Bbb N$ such that $|x_{3n+1}-L|<\epsilon$ whenever $n\ge m_\epsilon^{(1)}$.
What happens if you take $n\ge\max\{m_\epsilon^{(0)},m_\epsilon^{(1)}\}$? Then $|x_{3n}-L|<\epsilon$ and $|x_{3n+1}-L|<\epsilon$. Every integer is of one of the forms $3n$, $3n+1$, and $3n+2$. If you could also get $|x_{3n+2}-L|<\epsilon$ for all $n\ge\text{ something}$, you could get $|x_n-L|<\epsilon$ for a whole tail of the sequence $\langle x_n:n\in\Bbb N\rangle$. Look again at the last two paragraphs, and see if you can see how to ensure that $|x_{3n}-L|<\epsilon$, $|x_{3n+1}-L|<\epsilon$, and $|x_{3n_2}-L|<\epsilon$ for all $n$ from some point on.