# 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?

• 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? – Xander Henderson Dec 19 '18 at 4:13
• 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 – Jess Savoie Dec 19 '18 at 4:16

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.

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.

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.