Exercise 2.6.4

Question

Let $\left(a_{n}\right)$ and $\left(b_{n}\right)$ be Cauchy sequences. Decide whether each of the following sequences is a Cauchy sequence, justifying each conclusion.
  \begin{enumerate}[label=(\alph*)] 
  \item $c_{n}=\left|a_{n}-b_{n}\right|$
  \item $c_{n}=(-1)^{n} a_{n}$
  \item $c_{n}=\left[\left[a_{n}\right]\right]$, where $[[x]]$ refers to the greatest integer less than or equal to $x$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Yes. Note that by the Triangle Inequality, $$
    |a_n - a_m| + |b_m - b_n| + |a_m - b_m|
    \geq |a_n - b_n| \Rightarrow |a_n - a_m| + |b_m - b_n| \geq |a_n - b_n| - |b_m - a_m|
  $$
  and $$
    |a_m - a_n| + |b_n - b_m| + |a_n - b_m|
    \geq |a_m - b_m| \Rightarrow |a_n - a_m| + |b_m - b_n| \geq |a_m - b_m| - |b_n - a_n|
  $$
  therefore
  $$ |c_n - c_m| = \big| |a_n - b_n| - |a_m - b_m| \big | \leq |a_n - a_m| + |b_m - b_n| < \epsilon/2 + \epsilon/2 = \epsilon$$

\item No, if $a_n = 1$ then $(-1)^na_n$ diverges, and thus is not Cauchy.
  \item No, if $a_n = 1 - (-1)^n/n$ then $[[a_n]]$ fluctuates between $0$ and $1$ and so cannot be Cauchy.
   \end{enumerate}

Exercise 2.6.4

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Exercise 2.6.4

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