Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 52


Converges to $-2$

Work Step by Step

Given:$a_n=n-\sqrt {n+1}\sqrt {n+3}$ $a_n=n-\sqrt {(n+1)(n+3)}$ $a_n=n-\sqrt {n^{2}+4n+3}$ Multiply numerator and denominator by the conjugate. $n-\sqrt {n^{2}+4n+3}\times \frac{n+\sqrt {n^{2}+4n+3} }{n+\sqrt {n^{2}+4n+3}}=\frac{-4n-3}{n+\sqrt {n^{2}+4n+3}}$ $=\frac{-4-\frac{3}{n}}{1+\sqrt {1++\frac{4}{n}+\frac{3}{n}}}$ Find the limit as $n \to \infty $. $\lim\limits_{n \to \infty}\frac{-4-\frac{3}{n}}{1+\sqrt {1++\frac{4}{n}+\frac{3}{n}}}=\frac{-4-0}{1+\sqrt {1+0+0}}$ $=\frac{-4}{2}$ $=-2$ Hence, the sequence converges to $-2$.
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