A textbook problem is asking me to show that if $\lim_{n\to\infty}a_{2n}=L$ and $\lim_{n\to\infty}a_{2n+1}=L$, then $\{a_n\}$ is convergent and $\lim_{n\to\infty}a_{n}=L$. Here's my attempt:
It was assumed that $\lim_{n\to\infty}a_{2n}=L$ and $\lim_{n\to\infty}a_{2n+1}=L$, so given any $\varepsilon >0$, there exist $N_1\in\mathbb{N}$ and $N_2\in\mathbb{N}$ such that
$$2n>N_1\Rightarrow |a_{2n}-L|<\varepsilon$$
and
$$2n+1>N_2\Rightarrow |a_{2n+1}-L|<\varepsilon$$
Let $N=\max\{N_1,N_2\}$, and consider an arbitrary $n\in\mathbb{N}$. If $n>N$, then $n>N_1$ and $n>N_2$.
Clearly $n$ is either odd or even. If $n$ is odd, then $n=2k+1$ for some $k\in\mathbb{N}$, so $n>N_1$ and $n>N_2$ implies that $n=2k+1>N_2$, and consequently $|a_{n}-L|=|a_{2k+1}-L|<\varepsilon$. Similarly, $n$ being even implies that $|a_{n}-L|=|a_{2k}-L|<\varepsilon$ for some $k\in\mathbb{N}$. In either case, $|a_n-L|<\varepsilon$.
$n$ was picked arbitarily, so the preceding argument applies to all natural numbers. It follows that $\lim_{n\to\infty}a_n=L$
I'm a bit skeptical about this argument. Specifically, from the assumption that $\lim_{n\to\infty}a_{2n}=L$ and $\lim_{n\to\infty}a_{2n+1}=L$, is it valid to write $2n>N_1$ and $2n+1>N_2$ instead of $n>N_1$ and $n>N_2$?
Update: I found a way to justify my reasoning! You see, even if we use $n>N_1$ and $n>N_2$ instead of $2n>N_1$ and $2n+1>N_2$, it is still the case that $2n+1>n$ and $2n\geq n$ for all $n\geq 0$, so we still end up with $2n>N_1$ and $2n+1>N_2$. I'm so happy right now! 🙂
Best Answer
$$\exists N_1\ni\forall n>N_1, |a_{2n}-L|<\epsilon$$ $$\exists N_2\ni\forall n>N_2, |a_{2n+1}-L|<\epsilon$$ Take $N=2\max\{N_1,N_2\}$. Then for all $n>N$,
if $n=2k$, then $k> N_1$ and $|a_n-L|=|a_{2k}-L|<\epsilon$, or
if $n=2k+1$, then $k> N_2$, and $|a_n-L|=|a_{2k+1}-L|<\epsilon$.
Does this sound convincing?