Let $(a_n)$ and $(b_n)$ be convergent sequences. When is the sequence $(a_1,b_1,a_2,b_2,...)$ convergent and what is it's limit? I intuitively know that the sequence $(a_1,b_1,a_2,b_2,...)$ will only converge when the limits of $(a_n)$ and $(b_n)$ are equal, but how to I go about proving that? Can I use the uniqueness of limits? Do I need to prove the new sequence converges first?
 A: A Sequence converges to limit $L$ iff every subsequence converges to the same limit $L$.
$\{x_n\}=\{a_1,b_1,a_2,b_2,\cdots\}$ then $x_n$ converges iff $x_{2n-1}$ and $x_{2n}$ converge to the same limit.
A: Your sequence $(x_n)_n = (a_1,b_1,a_2,b_2, \ldots)$ converges if and only if $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n$.
Indeed, if $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = L$, then pick $\varepsilon > 0$. There exist $n_1, n_2 \in \mathbb{N}$  such that $$n \ge n_1 \implies |a_n - L| < \varepsilon$$
$$n \ge n_2 \implies |b_n - L| < \varepsilon$$
Then for $n_0 := 2\max\{n_1, n_2\}$ we have
$$n \ge n_0 \implies |x_n - L| = \begin{cases} |a_{k} - L|, \text{ if } n = 2k-1 \\ |b_k - L|, \text{ if } n = 2k\end{cases}$$
which is $< \varepsilon$ because in either case $k \ge n_1, n_2$. Therefore $\lim_{n\to\infty} x_n = L$.
Conversely, assume $\lim_{n\to\infty} a_n = L_1 \ne L_2 = \lim_{n\to\infty} b_n$ and $\lim_{n\to\infty} x_n = L$.
For $\varepsilon = |L_1 - L_2|$ there exists $n_0 \in \mathbb{N}$ such that $$n \ge n_0 \implies |x_n - L| < \frac\varepsilon4$$
However, $\exists n_1, n_2 \in \mathbb{N}$ such that 
$$n \ge n_1 \implies |a_n - L| < \frac\varepsilon4$$
$$n \ge n_2 \implies |b_n - L| < \frac\varepsilon4$$
Then for $k \ge \max\left\{n_0, n_1, n_2\right\}$ we have
\begin{align}
|L_1 - L_2| &\le |L_1 - x_{2k-1}| + |x_{2k-1} - L| + |L - x_{2k}| + |x_{2k} - L_2| \\
&= |L_1 - a_k| + |x_{2k-1} - L| + |L - x_{2k}| + |b_k - L_2|\\
&< \frac\varepsilon4 + \frac\varepsilon4 + \frac\varepsilon4 + \frac\varepsilon4\\
&= \varepsilon\\
&= |L_1 - L_2|
\end{align}
which is a contradiction. Hence $(x_n)_n$ doesn't converge.
A: On the one hand, if $a_n\to A$ and $b_n\to A$ when $n\to \infty$, then it's easy to prove that the sequence $\{a_1,b_1,a_2,b_2,\cdots\}$ converges to $A$.
On the other hand, if the limits of $\{a_n\}$ and $\{b_n\}$ distinct, then the sequence $\{a_1,b_1,a_2,b_2,\cdots\}$ cannot converge. Otherwise, it will have two subsequences which  converge to different limits, this is impossible.
