Let ($x_n$) and ($y_n$) be given, and define ($z_n$) to be the "shuffled" sequence $(x_1, y_1, x_2, y_2, x_3, y_3,...x_n, y_n)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are both convergent with $\lim x_n = \lim y_n$.

I know I need to prove both direction of the implication.

The first direction being if $(z_n)$ is convergent then so are $(x_n)$ and $(y_n)$ with $\lim x_n=\lim y_n$. I said let $(z_n)\rightarrow c$.

So since ($z_n$) converges, I know that given any positive $\varepsilon$, there exists N such that if $$n\geq N, z_n \in (c-\varepsilon, c+\varepsilon) \implies |x_n |<|c|+\varepsilon$$ for all $n\geq N$.

I'm stuck at the next step in seeing how this leads to proving that $(x_n)$ and $(y_n)$ converge. I'm inclined to say something like:

Since the elements of $(x_n)$ are also elements of $(z_n)$, then there exists $N_1$$\in$$\mathbb{N}$ such that $$n\geq N_1 \implies x_n \in(c-\varepsilon, c+\varepsilon)$$

And from there I'm stuck. Any helpful hints on what direction to go would be appreciated!

  • $\begingroup$ Then as $y_n$ converges, there exists a $N_2 \in \mathbb{N}$. $\endgroup$
    – Hetebrij
    Aug 30 '15 at 21:59
  • $\begingroup$ If a sequence converges to some limit, then so do all of its subsequences, and vice versa. $\endgroup$
    – Vim
    Aug 31 '15 at 1:54

Hint: A subsequence of a convergent sequence is convergent.

Conversely, setting $z_{2n - 1} = x_n $ and $z_{2n} = y_n$. As $\lim x_n = \lim y_n = a$, for any given $\varepsilon > 0$ there exists $n_1, n_2 \in \mathbb N $ such that

$$n > n_1 \implies |x_n - a| < \varepsilon \\ n > n_2 \implies |x_n - a| < \varepsilon$$

Consider $n_0 = \max \{2n_1 - 1, 2n_2\}$.

  • $\begingroup$ Then for all n$>$$n_o$, $\mid$$z_n$-a$\mid$$<$$\epsilon$, so $z_n$ converges to a ? $\endgroup$
    – Laura
    Aug 30 '15 at 22:21
  • $\begingroup$ That's correct. $\endgroup$ Aug 30 '15 at 22:22

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