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Is it true that if $x_n$ converges and $y_n$ is bounded, then $x_ny_n$ converges?

$x_n$ is said to be bounded if and only if it is bounded both above and below.

I believe this to be false. My attempt at a counterexample:

Let $x_n=(-1)^n(1-\frac{1}{n})$ and $y_n=(-1)^{3n}+2$

Then, $x_n$ converges to $1$ and $y_n$ will be bounded above at $3$ and bounded below at $1$

Thus, $x_ny_n$ will produce the sequence: $0,\frac{3}{2},-\frac{2}{3},\frac{9}{4}...$ that does not converge.

Is this an appropriate counterexample?

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2 Answers

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Following GitGud comment: A general set of examples will consist of $x_n\equiv c$ where $c\neq 0 $ is a constant and $y_n$ as any bounded non-convergent sequence

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If you are looking for non-constant sequences. You can try $\left\{(-1)^n \right\}$ and any non-constant sequence that has a non-zero limit.

For example $\left\{1+ \frac{1}{n}\right\}$. Then, $\left\{(-1)^{2n} (1+ \frac{1}{2n}) \right\}$ converges to $1$ whereas $\left\{(-1)^{2n-1} (1+ \frac{1}{2n-1}) \right\}$ converges to $-1$.

The trick was to produce two subsequences that have different limits.

It is interesting to note that if $\left\{x_n\right\}$ is a sequence that converges to $0$ and $\left\{y_n\right\}$ is bounded then $\left\{x_ny_n\right\}$ converges to $0$ as well.

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