Answer
$0$
Work Step by Step
We have
$$\lim_{n\to \infty}d_n=\lim_{n\to \infty}\ln(n^2+4)-\ln(n^2-1)\\
=\lim_{n\to \infty}\ln\frac{n^2+4}{n^2-1}=\ln 1=0.$$
Hence, by Theorem 1, the sequence $d_n$ converges to $0$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 538: 48
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