Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 13

Answer

(a) the $n$th term is ${a_n} = \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^3}}}$ for $n=1,2,3,....$ (b) the $n$th term is ${a_n} = \frac{{n + 1}}{{n + 5}}$ for $n=1,2,3,....$

Work Step by Step

(a) The sign of the numerator is alternating. If $n$ starts from 1, the general term of the numerator is ${\left( { - 1} \right)^{n + 1}}$. The general term of the denominator is $n^3$. Thus, the $n$th term is ${a_n} = \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^3}}}$ for $n=1,2,3,....$ (b) If $n$ starts from 1, the numerator is always one more than the index, whereas the denominator is five more than the index. So, the $n$th term is ${a_n} = \frac{{n + 1}}{{n + 5}}$ for $n=1,2,3,....$
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