The statement given is:

"If the sequence $\{x_n \cdot y_n\}_{n=1}^\infty$ converges to a limit $z^*$ and the sequence $\{x_n\}_{n=1}^\infty$ converges to the limit $x^*$, then there exists $y^*$ such that the sequence $\{y_n\}_{n=1}^\infty$ converges to the limit $y^*$."

To determine the truth of this statement, let's analyze it using the definitions and properties of sequences and limits.

1. If the statement is true, provide a proof.

Assume that:

1. $\{x_n \cdot y_n\}_{n=1}^\infty \to z^*$ (i.e., the product sequence converges to $z^*$).
2. $\{x_n\}_{n=1}^\infty \to x^*$ (i.e., the sequence $\{x_n\}$ converges to $x^*$).

We want to show that under these assumptions, $\{y_n\}$ must converge to some limit $y^*$.

By the definition of the limit of a product of sequences, if $x_n \cdot y_n \to z^*$ and $x_n \to x^*$, then it would follow that $x^* \cdot y^* = z^*$ if $\{y_n\}$ converges to $y^*$.

We can rewrite this convergence assumption as:

$\lim_{n \to \infty} x_n \cdot y_n = \lim_{n \to \infty} x_n \cdot \lim_{n \to \infty} y_n = x^* \cdot y^* = z^*$

However, this relation only implies that $x^* \cdot y^* = z^*$ if both limits exist, which is not guaranteed unless $x^* \neq 0$.

2. If the statement is false, prove that it is false.

The statement is indeed false without additional conditions.

Counterexample: Consider the following sequences:

• Let $x_n = \frac{1}{n}$ (which converges to $x^* = 0$).
• Let $y_n = n$ (which diverges).

Then the product sequence $x_n \cdot y_n = \frac{1}{n} \cdot n = 1$ for all $n$. So, $\{x_n \cdot y_n\}$ converges to $z^* = 1$. However, $\{x_n\}$ converges to 0, and $\{y_n\}$ diverges. This shows that even though $\{x_n \cdot y_n\}$ and $\{x_n\}$ converge, $\{y_n\}$ does not necessarily converge.

3. If the statement is false, is there a general condition on $x^*$ that, if satisfied, makes the statement true? If yes, provide a proof.

Yes, a condition on $x^*$ that makes the statement true is $x^* \neq 0$.

Proof: Suppose $x^* \neq 0$. Since $\{x_n\}$ converges to $x^*$, for sufficiently large $n$, $x_n$ will be close to $x^*$ and hence non-zero. This allows us to define:

$y_n = \frac{x_n \cdot y_n}{x_n}$

As $n \to \infty$, since $x_n \cdot y_n \to z^*$ and $x_n \to x^*$, it follows that:

$y_n = \frac{x_n \cdot y_n}{x_n} \to \frac{z^*}{x^*} = y^*$

This shows that $y_n$ converges to $y^* = \frac{z^*}{x^*}$. Hence, if $x^* \neq 0$, then $\{y_n\}$ indeed converges to $y^*$.

Conclusion

The statement is false in general, but it becomes true under the condition that $x^* \neq 0$. In this case, $\{y_n\}$ converges to $y^* = \frac{z^*}{x^*}$.

Would you like more details or have any questions? Here are five related questions to consider:

1. What happens if both $x^*$ and $z^*$ are zero?
2. Can a similar condition be derived for other types of convergence (e.g., uniform convergence)?
3. How does the behavior of $\{y_n\}$ change if $x_n$ has oscillatory behavior?
4. Can we extend these results to complex sequences?
5. How does the convergence speed of $\{x_n\}$ affect the convergence of $\{y_n\}$?

Tip: When analyzing sequence convergence, always consider the behavior of each sequence individually and in combination with others, especially when dealing with products or quotients of sequences.