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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?

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  • $\begingroup$ If $a_n \to L$ and $b_n \to L$, then the sequence $(a_1, b_1, a_2, b_2, \dotsc)$ will converge to $L$. Otherwise, this last sequence will oscillate. Where is your doubt? $\endgroup$ – Xander Henderson Dec 19 '18 at 4:13
  • $\begingroup$ Mostly the order I need to define terms to form the proof. It all makes sense but can I just say it would oscillate, or do I need to state more first $\endgroup$ – Jess Savoie Dec 19 '18 at 4:16
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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.

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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.

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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.

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