Solve the problem.
-Using the Maclaurin series for $\ln ( 1 + x )$, obtain a series for $\ln \frac { 1 + x } { 1 - x }$.
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 2 n + 1 }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 2 n + 1 }$
C) $2 \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 2 n + 1 }$
D) $2 \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 2 n + 1 }$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\ln(1+x)$ is $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}$. For $\ln(1-x)$, replace $x$ with $-x$ to get $\sum_{n=1}^{\infty} \frac{x^n}{n}$. The series for $\ln\frac{1+x}{1-x}$ is the difference: $\ln(1+x) - \ln(1-x)$, which simplifies to $2\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$ by combining and rearranging terms, focusing on the odd powers of $x$.

Solve the Problem ln⁡(1+x)\ln ( 1 + x )ln(1+x) , Obtain a Series For

Solve the Problem $\ln ( 1 + x )$ , Obtain a Series For