We observe that $f^{\text{'}}(x)=\frac{-1}{(1+x^2)}$ and find the required series by integrating the power series for $\frac{-1}{(1+x^2)}.$

$arccot(x)={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{-1}{(1+x^2)}dx$

$={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}-(1-+x^4-+$dots$)dx$

$=c-x+,-\frac{x^5}{5}+$

$$

To find $C,$ we put $x=0$ and obtain $C=arccot(0)=\frac{}{2}.$ Therefore we have the following.

$=\frac{}{2}-x+$

$-\frac{x^5}{5}+$

$$

$=\frac{}{2}-{\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{x^{2n+1}}{2n+1}$

Since the radius of convergence of the series for $\frac{-1}{(1+x^2)}$ is $$ the radius of convergence of this series is also $$