Cauchy Sequence proof with bounds

Question

Let $A_1, A_2, A_3,\dots$ be a Cauchy sequence of decreasing positive real numbers.

For $n = 1, 2, 3, \dots$ let $B_n$ be a real number such that:

$\sqrt{A(n+2011)} \le B_n \le \sqrt{A_n}$

Prove that $B_1, B_2, B_3, \dots$ is a Cauchy sequence by checking the definition of Cauchy sequence.

Answer 1

First you need to show that if $\{a_n\}$ is Cauchy then $\{ \sqrt{a_n} \}$. However, this is not difficult since $|\sqrt{a_n} - \sqrt{a_m}| = \frac{|a_n - a_m| } {\sqrt{a_n} + \sqrt{a_m} } \leq \frac{1}{2}|a_n - a_m|$. So for any $\epsilon > 0$ select N sufficiently large then $\forall n,m \geq N$ you have that $|\sqrt{a_n} - \sqrt{a_m} | < \epsilon$.

Using your upper and lower bound for $\{b_n\}$ in terms of the square roots of $a_n$ you can easily get the result.

Consider $|b_n -b_m|$ now $b_n < \sqrt{a_n}$ and $b_m > \sqrt{a_{m+2011}}$ which means that $-\sqrt{a_{m+2011}} < -b_m$ hence you can conclude that $|b_n-b_m| < |\sqrt{a_n} - \sqrt{a_{m+2011}}| < \epsilon$.

Note: Shifting m by 2011 is irrelevant as $\{\sqrt{a_n} \}$ is Cauchy.