## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 41

#### Answer

$\lim\limits_{n \to \infty} a_n=\sqrt 2$ and {$a_n$} is convergent.

#### Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt{\dfrac{2n}{n+1}}$ $\lim\limits_{n \to \infty} \sqrt{\dfrac{2n}{n+1}}=\sqrt{ \lim\limits_{n \to \infty}\dfrac{2n}{n+1}}$ or, $=\sqrt{ \lim\limits_{n \to \infty}\dfrac{2}{1+\dfrac{1}{n}}}$ or, $=\sqrt 2$ Thus, $\lim\limits_{n \to \infty} a_n=\sqrt 2$ and {$a_n$} is convergent.

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