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Given the sequence $\{x_n\}$ such that $\lim_{n \to \infty}{x_{2n}} = a$ and $\lim_{n \to \infty}{x_{2n+1}} = a$, show that $\lim_{n \to \infty}{x_{n}} = a$

Can I do this by sub sequences?

I need some help to do this.

Any suggestion?

Curious
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  • o prove $\lim_{n\to \infty} a$ you need to show that, given any $\epsilon> 0$ they} f(x)=re exist N such that if n> N then $|x_n- a|< \infty$. Do that by treating n even and n odd separately. – user247327 Mar 25 '21 at 16:39

2 Answers2

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The following result may be useful:

A sequence $\{x_n\}$ converges to a limit $a$ if and only if every subsequence of $\{x_n\}$ has a further subsequence that converges to $a$.

See here.

Now, choose any subsequence, which I will just call it $L$, of $\{x_n\}$. Note that since $\{x_{2n}\}$ and $\{x_{2n+1}\}$ partitions $\{x_n\}$, then $L$ must contain infinitely many of that $\{x_{2n}\}$’s or $\{x_{2n+1}\}$.

In other words, $L$ contains a subsequence of either $\{x_{2n}\}$ or $\{x_{2n+1}\}$, which converges to $a$ either way, since $\{x_{2n}\}$ and $\{x_{2n+1}\}$ converges to $a$. So by the result, $\{x_{n}\}$ converges to $a$.

Soby
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Let us consider an $ε>0$. We have that $x_{2n}\to a$ so there exists $n_1\in\mathbb{N}$ so that for every $n\geq n_1$ we have $|x_{2n}-a|<ε$. Same goes for the sequence $(x_{2n+1})$, there exists $n_2\in\mathbb{N}$ so that for every $n\geq n_2$ we have $|x_{2n+1}-a|<ε$. Take $n_0=max\{2n_1,2n_2+1\}$. Now for every $n\geq n_0$ show that the relation $|x_n-a|<ε$ holds true.

Just_a_student
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