How was this Maclaurin expansion derived? For each $|x|<1,$
\begin{align} \left( \frac{\arctan(x)}{1-x} \right)&=\left( \sum^{\infty}_{k=0}x^k\right)\left(\sum^{\infty}_{j=0}\dfrac{(-1)^j x^{2j+1}}{2j+1}\right)\\&= \sum^{\infty}_{k=0}\left(\sum_{j\in D_k} \dfrac{ (-1)^{j} }{2j+1}\right)x^k,\;\text{where}\; D_k=\{j\in \Bbb{N}:0\leq j\leq (k-1)/2\}.\end{align}
HERE'S MY TRIAL
\begin{align} \left( \frac{\arctan(x)}{1-x} \right)&=\left( \sum^{\infty}_{k=0}x^k\right)\left(\sum^{\infty}_{j=0}\dfrac{(-1)^j x^{2j+1}}{2j+1}\right)\\&= \sum^{\infty}_{k=0}\left(\sum^{k}_{j=0} x^j\dfrac{ (-1)^{(k-j)} x^{2(k-j)+1}}{2(k-j)+1}\right),\;\text{for}\; k\in \Bbb{N}\\&\stackrel{\text{how?}}{=} \sum^{\infty}_{k=0}\left(\sum_{j\in D_k} \dfrac{ (-1)^{j} }{2j+1}\right)x^k.\end{align}
Best Answer
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 $$ \sum_{m=0}^\infty\sum_{j=0}^{\left\lfloor\frac{m-1}2\right\rfloor}\overbrace{\ \frac{(-1)^j}{2j+1}\ }^{b_j}x^m $$