Question

Task
If (a sub_n) and (b sub_n) are cauchy sequences, show directly that (a sub_n * b sub_n) is also a cauchy sequence.
Detailed explanation: NO
Specific requirements:
Prove

Answer 1

Question 1. If $(a_n)$ and $(b_n)$ are Cauchy sequences, show directly that $(a_n b_n)$ is also a Cauchy sequence.

Solution. First prove that each Cauchy sequence $(x_n)$ is bounded. Fix $\varepsilon > 0$. By the definition of Cauchy sequence there is $N \in \mathbb{N}$ such that $|x_n - x_m| < \varepsilon$ for all $m, n \geq N$. Then for all $n \geq N$ we have

$$\left| x _ {n} \right| = \left| x _ {n} - x _ {N} + x _ {N} \right| \leq \left| x _ {n} - x _ {N} \right| + \left| x _ {N} \right| < \varepsilon + \left| x _ {N} \right|.$$

Therefore, $|x_{n}| \leq \max \{|x_{1}|, \ldots, |x_{N - 1}|, |x_{N}| + \varepsilon\} < \infty$.

Now let $(a_{n})$ and $(b_{n})$ be Cauchy sequences. There are $A, B > 0$ such that $|a_{n}| < A$ and $|b_{n}| < B$ for all $n \in \mathbb{N}$, as shown above. Fix $\varepsilon > 0$ and find $N_{1}, N_{2} \in \mathbb{N}$ such that $|a_{n} - a_{m}| < \frac{\varepsilon}{2B}$ for all $m, n > N_{1}$ and $|b_{n} - b_{m}| < \frac{\varepsilon}{2A}$ for all $m, n > N_{2}$. Set $N = \max \{N_{1}, N_{2}\}$. Then for all $m, n > N$ we have

$$\begin{array}{l} \left| a _ {n} b _ {n} - a _ {m} b _ {m} \right| = \left| a _ {n} \left(b _ {n} - b _ {m}\right) + b _ {m} \left(a _ {n} - a _ {m}\right) \right| \\ \leq \left| a _ {n} \right| \cdot \left| b _ {n} - b _ {m} \right| + \left| b _ {m} \right| \cdot \left| a _ {n} - a _ {m} \right| \\ < A \cdot \frac {\varepsilon}{2 A} + B \cdot \frac {\varepsilon}{2 B} \\ = \frac {\varepsilon}{2} + \frac {\varepsilon}{2} \\ = \varepsilon . \\ \end{array}$$

Thus, $(a_{n}b_{n})$ is Cauchy sequence.