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> Let $(x\_n)$ and $(y\_n)$ be given sequences 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 convergent and $\lim x\_n=\lim y\_n$.
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Can someone prove this theorem for me? I am lost and all I have so far is:
So we can write that $(z\_{2n-1})=(x\_n)$ and $(z\_{2n})=(y\_n)$ for $n \in \mathbb N.$

$(\Longrightarrow)$ (Is this correct?)

Assume for $n \ge N\_1 $ $\Rightarrow$ $|x\_{n}-l|< \epsilon$ and assume for $n \ge N\_2 $ $\Rightarrow$ $|y\_{n}-l|< \epsilon$. Choose $M \ge max\{N\_1, N\_2\}$ and then for $n \ge M$ for $2n$ and $2n+1$ is we get $|z\_n -l| \lt \epsilon$.

($\Longleftarrow$) I don't have anything for this direction.

Note that this question [If $x\_n \to a$ and $x'\_n \to a$, then $\{x\_1, x'\_1, x\_2, x'\_2, ...\} \to a$](https://math.stackexchange.com/questions/1275913/if-x-n-to-a-and-x-n-to-a-then-x-1-x-1-x-2-x-2-to-a) only addresses the theorem in one direction so it isn't exactly a duplicate.