Compose a square-root series with an inverse tangent

$y^{\prime}=1/(1+x^2)$, so $(1+x^2)y^{\prime}=1$. Differentiating gives $(1+x^2)y^{\prime\prime}+2xy^{\prime}=0$. Differentiate again to obtain $(1+x^2)y^{\prime\prime\prime}+4xy^{\prime\prime}+2y^{\prime}=0$.

At zero, $y=0$, $y^{\prime}=1$, $y^{\prime\prime}=0$, $y^{\prime\prime\prime}=-2$. Hence $\tan^{-1}x=x-x^3/3+O(x^5)$ and $h=-1/3$.

Put $u=\tan^{-1}x=x-x^3/3+\cdots$ in $(1+u)^{1/2}=1+u/2-u^2/8+u^3/16+\cdots$. Through degree three, $u^2=x^2+O(x^4)$ and $u^3=x^3+O(x^5)$, giving

$\begin{aligned}&\sqrt{1+\tan^{-1}x}\\ & =1+\frac{x}{2}-\frac{x^2}{8}-\frac{5x^3}{48}\\ &{}+O(x^4).\end{aligned}$