Question

Q: Give an example of two divergence sequences X and Y such that:
(a) their sum X+Y converges, (b) their product XY converges

Answer 1

Answer on Question #64522 – Math – Real Analysis

Question

Give an example of two divergence sequences X and Y such that: (a) their sum X+Y converges, (b) their product XY converges

Solution

Let's consider the sequences $x_{n} = \begin{cases} 0, & n - odd \\ 1, & n - even \end{cases}$ and $y_{n} = \begin{cases} 1, & n - odd \\ 0, & n - even \end{cases}$ .

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

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

PART(a). Now we can consider the sum of those sequences: $x_{n} + y_{n} = \begin{cases} 1, & n - odd \\ 1, & n - even \end{cases} = 1 \xrightarrow[n \to \infty]{} 1$ , hence the sequence $(x_{n} + y_{n})$ is convergent.

PART(b). Now we can consider the product of those sequences: $x_{n} \cdot y_{n} = \begin{cases} 0, & n - odd \\ 0, & n - even \end{cases} = 0 \xrightarrow[n \to \infty]{} 0$ , hence the sequence $(x_{n} \cdot y_{n})$ is convergent.

Answer: $x_{n} = \begin{cases} 0, & n - odd \\ 1, & n - even \end{cases}, y_{n} = \begin{cases} 1, & n - odd \\ 0, & n - even \end{cases}.$

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