Proving convergence of a bizarre sequence

Question

This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?

Suppose that $\{a_n\}$ and $\{b_n\}$ are two convergent sequences that both converge to the same $L \in \mathbb{R}$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $\{c_n\}$ converges to $L$.

Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
$$\lim \sup a_n = \lim \inf a_n = \lim \sup b_n = \lim \inf b_n = L$$
Define $i_n = \inf\{c_k : k \geq n\}$ and $s_n = \sup\{c_k : k \geq n\}$. Then $\lim i_n = \lim \inf c_n$ and $\lim s_n = \lim \sup c_n$.

$L$ is a subsequential limit of $c_n$ and so $\lim \inf c_n \leq L \leq \lim \sup c_n$.

Suppose for contradiction that $\lim \inf c_n < L$. Then there exists an $n_0$ such that $L \neq i_{n_0}$ and $i_{n_0} < a_{n_0}$ or $i_{n_0} < b_{n_0}$, but this contradicts the construction of $\{a_n\}$ and $\{b_n\}$ because one or both would not converge at all or would converge to $i_{n_0}$. Thus, $\lim \inf c_n = L$.

A similar argument can be used to show that $\lim \sup c_n = L$ by replacing $i_{n_0}$ with $s_{n_0}$, $\lim \inf c_n$ with $\lim \sup c_n$ and switching the inequality signs in the previous paragraph.

Since $\lim \inf c_n = L = \lim \sup c_n$, $c_n$ converges to $\lim a_n = \lim b_n = L$.

Answer 1

I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.

For a simpler approach try this without invoking $\sup$ or $\inf$. If you need help see below:

Let $\epsilon>0 $ be given we need to show that $\exists N_c$ such that $\forall n>N_c$ we have $|c_n-L|<\epsilon$ but because we know $\{a_n\}$ converges we can select $N_a$ such that $\forall n>N_a$ we have $|a_n-L|<\epsilon$ and $\{b_n\}$ converges we can select $N_b$ such that $\forall n>N_b$ we have $|b_n-L|<\epsilon$. So we can just take $N_c=\max\{N_a,N_b\}$ and then we are guaranteed to be within $\epsilon$ of our target.