Sticky Summation (2)

Relevant wiki: Maclaurin Series

Let $S = \displaystyle \sum_{n=0}^\infty \frac {(-1)^n}{(n+2)4^n}$ . Now consider,

$\begin{aligned} \ln (1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \cdots & \small \color{#3D99F6} \text{Maclaurin series, for }-1 < x \le 1\\ & = x - x^2 \left(\frac 12 - \frac x3 + \frac {x^2}4 + \cdots \right) \\ & = x - x^2 \sum_{n=0}^\infty \frac {(-1)^n x^n}{n+2} & \small \color{#3D99F6} \text{Putting }x = \frac 14 \\ \ln \left(1+\frac 14\right) & = \frac 14 - \frac 1{16}S \\ \implies S & = 4 - 16 \ln \left(\frac 54\right) \end{aligned}$

$\implies a+b = 4+5 = \boxed{9}$