Answer to Question #177468 in Real Analysis for Vinay

Question

Give an example of a divergent sequence which has two convergent sequences. Justify

your claim.

Answer 1

Example of a divergent sequence which has two convergent sequences.

Let's consider the sequences $x_n = \begin{cases} 0 &\text{if } n-odd \\ 1 &\text{if } n-even \end{cases}$ and $y_n = \begin{cases} 1 &\text{if } n-odd \\ 0 &\text{if } n-even \end{cases}$

Since the subsequences $x_{2k}=1_{k\to\infty}\to1$ and $x_{2k-1}=0_{k\to\infty}\to0$ have different limit points, then the sequence $x_n = \begin{cases} 0 &\text{if } n-odd \\ 1 &\text{if } n-even \end{cases}$ is divergent.

Since the subsequences $y_{2k}=1_{k\to\infty}\to1$ and $y_{2k-1}=0_{k\to\infty}\to0$ have different limit points, then the sequence $y_n = \begin{cases} 1 &\text{if } n-odd \\ 0 &\text{if } n-even \end{cases}$ is divergent.

Now, we can consider the sum of those sequences:

$x_n+y_n = \begin{cases} 1 &\text{if } n-odd \\ 1 &\text{if } n-even \end{cases}= 1_{n\to \infty}\to1,$ Hence the sequence $(x_n+y_n)$ is convergent.

And also, we can consider the product of those sequences:

$x_n\times y_n = \begin{cases} 0 &\text{if } n-odd \\ 0 &\text{if } n-even \end{cases}= 1_{n\to \infty}\to1,$ Hence the sequence $(x_n\times y_n)$ is convergent.

Final Answer: $x_n = \begin{cases} 0 &\text{if } n-odd \\ 1 &\text{if } n-even \end{cases}$ , $y_n = \begin{cases} 1 &\text{if } n-odd \\ 0 &\text{if } n-even \end{cases}$