Maclaurin expansion of $\arctan(x)/(1 − x).$

The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is $$ \sum_{j=0}^{\left\lfloor\frac{m-1}2\right\rfloor}\overbrace{\vphantom{b}\ a_{m-2j-1}\ }^{\substack{\text{coefficient}\\\text{of $x^{m-2j-1}$}}}\overbrace{\ \ \ \ \ b_j\ \ \ \ \ }^{\substack{\text{coefficient}\\\text{of $x^{2j+1}$}}} $$ That is, $$ \sum_{k=0}^\infty a_kx^k\sum_{j=0}^\infty b_jx^{2j+1} =\sum_{m=0}^\infty\sum_{j=0}^{\left\lfloor\frac{m-1}2\right\rfloor}a_{m-2j-1}b_jx^m $$ Since $a_k=1$ for all $k\ge0$, we have $$ \sum_{m=0}^\infty\sum_{j=0}^{\left\lfloor\frac{m-1}2\right\rfloor}\overbrace{\ \frac{(-1)^j}{2j+1}\ }^{b_j}x^m $$

Corrected: Thanks to robjohn for the mentorship. Since \begin{align} \left( \sum^{\infty}_{k=0}a_k x^k\right)\left( \sum^{\infty}_{j=0}b_j x^j\right)&=\sum^{\infty}_{r=0}\left( \sum^{r}_{j=0}b_j a_{r-j} \right)x^{r},\;\text{where}\; r=k+j\;\text{and}\;r\in\Bbb{N},.\end{align} we have that,

\begin{align} \left( \frac{\arctan x }{1-x} \right)&=\left( \sum^{\infty}_{k=0}a_k x^k\right)\left( \sum^{\infty}_{j=0}b_j x^{2j+1}\right)\\&=\sum^{\infty}_{\gamma=0}\left( \sum^{\frac{\gamma-1}{2}}_{j=0}b_j a_{\gamma-2j-1} \right)x^{\gamma},\;\text{where}\; \gamma=k+2j+1\;\text{and}\;\gamma\in\Bbb{N},\\&=\sum^{\infty}_{\gamma=0}\left( \sum^{\frac{\gamma-1}{2}}_{j=0}\dfrac{ (-1)^{j} }{2j+1} \right)x^{\gamma},\\&= \sum^{\infty}_{\gamma=0}\left(\sum_{j\in D_\gamma} \dfrac{ (-1)^{j} }{2j+1}\right)x^\gamma.\end{align} where $D_\gamma=\{j\in \Bbb{N}:0\leq j\leq (\gamma-1)/2\},\;a_{\gamma-2j-1} =1,\,j\in D_\gamma ,\;b_j= (-1)^{j} /(2j+1).$