Question: Let $(x_n)$ be a Cauchy sequence and $(y_n)$ a bounded sequence. Prove that $(x_n^2y_n)$ has a convergent subsequence.

Proof (can someone verify it?):

Since $(x_n)$ is Cauchy, it is convergent. Let $(x_n)$ converge to some $x \in \mathbb{R}$.

Since $(y_n)$ is bounded, lim sup $y_n$ exists, and is a real number $\gamma$. Then, there exists a subsequence $(y_{n_k})$ converging to $\gamma$. Also, $(x_{n_k})$ converges to $x$. Using limit theorems, it can be inferred that $(x_{n_k}^2y_{n_k})$ is convergent, with limit $x^2\gamma$.

  • $\begingroup$ looks good to me $\endgroup$ – user113529 May 16 '14 at 14:03
  • 3
    $\begingroup$ Yes, this is correct, but you need to state that you're working in $\mathbb{R}$. $\endgroup$ – Edvard Fagerholm May 16 '14 at 14:03

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