## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 32

#### Answer

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

#### Work Step by Step

Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n+3}{n^2+5n+6}$ or, $\lim\limits_{n \to \infty} \dfrac{n+3}{n^2+5n+6}= \lim\limits_{n \to \infty} \dfrac{n+3}{(n+3)(n+2)}$ Now, apply Sandwich Theorem which states that $\lim\limits_{n \to 0} \dfrac{1}{n}=0$ Thus, $\lim\limits_{n \to \infty} \dfrac{1}{n+2}=0$ Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.

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