[Math] Prove shuffled sequence $\{x_i, y_i\}$ converges $\iff \lim x_n = \lim y_n$ (Abbott p 49 q2.3.5)

proof-verificationreal-analysissequences-and-series

Let $(x_{n})$ and $(y_{n})$ be given. Define $(z_{n})$ to be the shuffled sequence $(x_{1}.y_{1},\ x_{2},\ y_{2},\ x_{3},\ldots,x_{n}, y_{n},\ldots)$ . Prove that $(z_{n})$ is conv ergent $\iff (x_{n})$ and $(y_{n})$ are both convergent $\backslash $ with $ \lim x_{n}=\lim y_{n}$.

$(\Rightarrow)$ Let $\epsilon >0$ be arbitrary. Let's call the limit that $(z_{n})$ converges to $L$. Then we must find a natural number $N$ such that when $n \geq N$, then $|y_{\gamma 1}-L|<\epsilon$. Because $(z_{n})\rightarrow L$, we can pick $N_1$ so that $| z_{n}-L|<\epsilon$ for all $n\geq N_1$. Because $y_{n}=z_{2N}$, it certainly follows that $|y_{n}-L|<\epsilon$ whenever $n\geq N_1.$ A similar argument holds for the $(x_{n})$ sequence, because $x_n = z_{2N – 1}$.

1. Why did solution capitalize in $y_{n}=z_{2\color{red}{N}}$? Is it at fault to write lowercase : $y_{n}=z_{2\color{red}{n}}$ for all $n \in N$? By the agency of the definition, $y_n := z_{even}$ and $x_n := z_{odd}$.

$(<=)$ Let $\epsilon>0$ be arbitrary. Again, let $L$ be the common limit of $(x_{N})$ and $(y_{n})$ . We need to show that there exists an $N$ such that when $n\geq N$, then $|z_{N}-L|<\epsilon$. Choose $N_1$ so $|x_{n}-L|<\epsilon$ for all $n\geq N_{1}$, and choobe $N_{2}$ such that $|y_{n}-L|<\epsilon$ for all $n\geq N_{2}$. Finally, let $N =\max\{\color{magenta}{2}N_{1},\color{magenta}{2}N_{2}\}$. Then from the construction of the sequence $(z_{n})$, it follows $|z_{n}-L|<\epsilon$ whenever $n \geq N$.

2. When I first did this, I picked $N =\max\{N_{1},N_{2}\}$. Is this faultless? Why do we need the $\color{magenta}{2}$?
3. Please explain 'from the construction of the sequence'? I don't understand 'construction' here. How does convergence of $\{z_n\}$ follow?

@HagenvonEitzen Sorry…I don't understand your answers. What does 'yes' mean for 1. ?
For 2., is it at fault to exclude the $\color{magenta}{2}$ ? Why use $\color{magenta}{2}$?

Best Answer

  1. Yes

  2. You want to show $|z_n-L|<\epsilon$ for $n>N$. By the way $z_n$ is defined, you have either $z_n=x_k$ with $n=2k-1$, or $z_n=y_k$ with $n=2k$. So to make the conclusion, you'd want $|x_k-L|<\epsilon$ and $|y_k-L|<\epsilon$, respectively. You can do so if $k>N_1$ and $k>N_2$, respectively, i.e. $k>\max\{N_1,N_2\}$. If $n>2\max\{N_1,N_2\}$, this works.

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