Proving $\lim\limits_{n\to\infty}x_n=a$ iff $\lim\limits_{n\to\infty}x_{2n}=a=\lim\limits_{n\to\infty}x_{2n+1}$

Question

Prove $$\lim\limits_{n\to\infty}x_n=a\quad\iff\quad\lim\limits_{n\to\infty}x_{2n}=a=\lim\limits_{n\to\infty}x_{2n+1}$$

I try to prove this using $\varepsilon-\delta$ definition, there exist N, when n > N ,$|a_n - a|< \varepsilon$

how to prove it from right side to left side?

There is my proof: $\lim\limits_{n\to\infty}x_n=a\iff\lim\limits_{n\to\infty}x_{2n}=a=\lim\limits_{n\to\infty}x_{2n+1}$ $\because\lim\limits_{n\to\infty}x_n=a,\exists N,n>N,|a_n-a|<\varepsilon\implies N=\frac{2}{\varepsilon},|a_{2n}-a|<\varepsilon,N=\frac{2}{\varepsilon}+1,|a_{2n+1}-a|<\varepsilon$ $\therefore\lim\limits_{n\to\infty}x_n=a\iff\lim\limits_{n\to\infty}x_{2n}=a=\lim\limits_{n\to\infty}x_{2n+1}$

$N=max\{2N,2(N+1)\},|a_{2n}-a|<\varepsilon,|a_{2n+1}-a|<\varepsilon\implies |a_n-a|<\varepsilon$

If there is some sub-sequences,a part sub-sequences if converage, can I get a output the the sequence is converage.

Try to prove a sequence all sub-squence is converage that mother sequence is converage can solve it. Right ?

Answer 1

Going from left to right:

Assume $\lim_{n\to \infty}x_n=a$. Then $\forall\epsilon>0$, $\exists N$ such that whenever $n>N$ then $|x_n-a|<\epsilon.$

Then with the same $N$ we also have $|x_{2n}-a|<\epsilon$ $|x_{2n+1}-a|<\epsilon$ whenever $n>N$ simply because $2n\geq n$ and $2n+1\geq n$.

Going from right to left. Let $\epsilon>0$ then there exists $N_1$ and $N_2$ such that

$|x_{2n}-a|<\epsilon$ when $n>N_1$ and $|x_{2n+1}-a|<\epsilon$ when $n>N_2$.

Then we can pick $N=2\times max(N_1,N_2)+1$ then whenever $n>N$ we have $2$ cases, either $n$ is odd or $n$ is even. Each case satisfies $|x_n-a|<\epsilon$ simply due to our choice of N.