## 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: 26

#### Answer

Radius of convergence is $1$ $\int \frac{t}{1+t^{3}}dt=c+\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{t}^{3n+2}}{3n+2}$

#### 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, $\frac{t}{1+t^{3}}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(t)(-t^{3})^{n}=\Sigma_{n=0}^{\infty}(-1)^{3}{t}^{3n+1}$ This is the power series representation of $f(t)$. we know that the power series converges when $r=|t|\lt 1$ Radius of convergence is $1$ $\int \frac{t}{1+t^{3}}dt=\Sigma_{n=0}^{\infty}(-1)^{3}{t}^{3n+1} =c+\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{t}^{3n+2}}{3n+2}$

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