Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.7 Taylor Series - Preliminary Questions - Page 588: 4


$$f (x)=4+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$

Work Step by Step

Given $$f^{\prime}(x)=\sum_{n=1}^{\infty} \frac{(x-3)^{n}}{n}$$ By integration, we get $$f (x)=C+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$ Since $f(3)=4 $, then $C=4$, hence $$f (x)=4+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$
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