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: 55

Answer

converges to $1$

Work Step by Step

As we know that when $x \gt 0$, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt[n] {10n}$ This implies that $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt[n] {10n}$ and $a_n= \lim\limits_{n \to \infty} 10^{1/n} \cdot n^{1/n}=1 \cdot 1=1$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.
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