Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 570: 40

Answer

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{-1}{2})^n$ Since, we know that $|\dfrac{-1}{2}|=\dfrac{-1}{2} \lt 1$.This means that $ \lim\limits_{n \to \infty} (\dfrac{-1}{2})^n=0$ Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
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