Provide an appropriate response.
-Use the fact that $\tan ^ { - 1 } x = \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$ for $| x |

Question

Provide an appropriate response.
-Use the fact that $\tan ^ { - 1 } x = \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$ for $| x | < 1$ to find the series for $\cot ^ { - 1 } x$.
A) $\frac { \pi } { 2 } - \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$
B) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$
C) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$
D) $\frac { \pi } { 2 } - \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$

Quizplus · Accepted Answer

Using the identity $\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x$, we have
\begin{align*}
\cot^{-1}x &= \frac{\pi}{2} - \tan^{-1} x \
&= \frac{\pi}{2} - \sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{2n-1} 
\end{align*}
where we used the given identity for $\tan^{-1}x$. Rearranging the terms gives the series for $\cot^{-1}x$:
$$\cot^{-1}x = \sum_{n=1}^\infty \frac{(-1)^{n} x^{2n-1}}{2n-1}$$ 
which matches choice A.

Provide an Appropriate Response tan⁡−1x=∑n=1∞(−1)n−1x2n−1(2n−1)\tan ^ { - 1 } x = \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }tan−1x=∑n=1∞​(2n−1)(−1)n−1x2n−1​

Provide an Appropriate Response $\tan ^ { - 1 } x = \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n - 1 } } { ( 2 n - 1 ) }$