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 $$ \sum_{m=0}^\infty\sum_{j=0}^{\left\lfloor\frac{m-1}2\right\rfloor}\overbrace{\ \frac{(-1)^j}{2j+1}\ }^{b_j}x^m $$