Chapter 9 - Infinite Series - 9.8 Exercises - Page 654: 13

$\left( { - 1,1} \right]$

$$\eqalign{ & \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^n}}}{n}} \cr & {\text{Let }}{u_n} = \frac{{{{\left( { - 1} \right)}^n}{x^n}}}{n} \cr & {\text{Using the ratio test}} \cr & \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{{\left( { - 1} \right)}^{n + 1}}{x^{n + 1}}}}{{n + 1}}}}{{\frac{{{{\left( { - 1} \right)}^n}{x^n}}}{n}}}} \right| \cr & {\text{Simplifying}} \cr & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}{x^{n + 1}}}}{{n + 1}} \cdot \frac{n}{{{{\left( { - 1} \right)}^n}{x^n}}}} \right| \cr & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^n}\left( { - 1} \right){x^n} \cdot x}}{{n + 1}} \cdot \frac{n}{{{{\left( { - 1} \right)}^n}{x^n}}}} \right| \cr & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( { - 1} \right) \cdot x}}{{n + 1}} \cdot n} \right| \cr & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{nx}}{{n + 1}}} \right| \cr & = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\frac{n}{{n + 1}}} \right| \cr & {\text{Evaluating the limit}} \cr & = \left| x \right|\left( 1 \right) \cr & = \left| x \right| \cr & {\text{By the Ratio Test}},{\text{ the series converges for }} \cr & \left| x \right| < 1 \cr & - 1 < x < 1 \cr & \cr & {\text{At the endpoints}}{\text{, we have}} \cr & x = - 1 \to \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( { - 1} \right)}^n}}}{n}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left[ {\left( { - 1} \right)\left( { - 1} \right)} \right]}^n}}}{n} = } \sum\limits_{n = 1}^\infty {\frac{1}{n},{\text{ diverges}}} \cr & x = 1 \to \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( 1 \right)}^n}}}{n}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n},{\text{ This is an alternating series}}} , \cr & {\text{then at }}x = 1{\text{ the series converges}}{\text{.}} \cr & {\text{The interval of convergence is}} \cr & \left( { - 1,1} \right] \cr} $$