Answer
(a) $M = \frac{5}{{{{\log }_{10}}3}}$
(b) $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$
Work Step by Step
(a) We have
$\left| {{b_n}} \right| \le {10^{ - 5}}$.
Since ${b_n} = {\left( {\frac{1}{3}} \right)^n}$, so
$\frac{1}{{{3^n}}} \le {10^{ - 5}}$,
${10^5} \le {3^n}$,
$5 \le n{\log _{10}}3$.
$n \ge \frac{5}{{{{\log }_{10}}3}}$.
We can choose $M = \frac{5}{{{{\log }_{10}}3}}$.
(b) From part (a) we have
$\left| {{b_n} - 0} \right| \le {10^{ - 5}}$ for $n \ge \frac{5}{{{{\log }_{10}}3}}$.
Since there is a number $M = \frac{5}{{{{\log }_{10}}3}}$ such that the above inequality is valid. By the limit definition, this proves that $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$.
