Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 538: 71

Answer

(b)

Work Step by Step

According to the limit definition: We write $\mathop {\lim }\limits_{n \to \infty } {a_n} = L$ if, for every $\varepsilon > 0$, there is a number $M$ such that $\left| {{a_n} - L} \right| < \varepsilon $ for all $n>M$. Thus, $\left| {{a_n} - L} \right| < \varepsilon $, $ - \varepsilon < {a_n} - L < \varepsilon $, $L - \varepsilon < {a_n} < L + \varepsilon $, ${\ \ \ }$ for all $n>M$. The last inequality is equivalent to saying the interval $\left( {L - \varepsilon ,L + \varepsilon } \right)$ contains all elements of the sequence $\left\{ {{a_n}} \right\}$ for $n>M$, except possibly some finite number of elements ${a_1},{a_2},{a_3},...,{a_M}$. In other words, stated in (b): for every $\varepsilon > 0$, the interval $\left( {L - \varepsilon ,L + \varepsilon } \right)$ contains all but at most finitely many elements of the sequence $\left\{ {{a_n}} \right\}$.
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