## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 798: 25

#### Answer

Radius of convergence is $1$ $\int \frac{t}{1-t^{8}}dt=c+\Sigma_{n=0}^{\infty}\dfrac{{t}^{8n+2}}{{t}^{8n+2}}$

#### Work Step by Step

Sum of geometric series with initial term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Here, $\frac{t}{1-t^{8}}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(t)(t^{8})=\Sigma_{n=0}^{\infty}{t}^{8n+1}$ This is the power series representation of $f(t)$. we know that the power series converges when $r=|t^{8}|\lt 1$ Interval of convergence is $(-1,1)$ Radius of convergence is $1$ $\int \frac{t}{1-t^{8}}dt=\int[ \Sigma_{n=0}^{\infty}{t}^{8n+1}]dt =c+\Sigma_{n=0}^{\infty}\dfrac{{t}^{8n+2}}{{t}^{8n+2}}$

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