$P$ is for $P$ ascal

Note that $S_n$ can be rewritten as:

$\begin{aligned} S_n & = \sum_{k=1}^n \frac {R_k}{R_{2n+1-k}} & \small \color{#3D99F6} \text{By binomial theorem: }R_m = \sum_{k=0}^{m-1} \binom {n-1}k = 2^{m-1} \\ & = \sum_{k=1}^n \frac {2^{k-1}}{2^{2n-k}} & \small \color{#3D99F6} \text{Multiply up and down by }2^{k+1} \\ & = \sum_{k=1}^n \frac {2^{2k}}{2^{2n+1}} \\ & = \frac 1{2^{2n+1}} \color{#3D99F6} \sum_{k=1}^n 4^k & \small \color{#3D99F6} \text{Sum of a geometric progression} \\ & = \frac 1{2(4^n)} \color{#3D99F6} \left(\frac {4(4^n-1)}{4-1} \right) \\ & = \frac 23 \left(1 - \frac 1{4^n} \right) \end{aligned}$

$\displaystyle \implies \lim_{n \to \infty} S_n = \dfrac 23$ and $a+b = 2+3 = \boxed 5$ .

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Reference: Binomial theorem

$S_n= \frac{1}{R_{2n}}+\frac{1+1}{R_{2n-1}}+\frac{1+2+1}{R_{2n-2}}+\frac{\dots}{\dots}+\frac{R_{n-1}}{R_{n+2}}+\frac{R_n}{R_{n+1}}$ $S_n= \frac{R_n}{R_{n+1}}+\frac{R_{n-1}}{R_{n+2}}+\frac{\dots}{\dots}+\frac{1+2+1}{R_{2n-2}}+\frac{1+1}{R_{2n-1}}+\frac{1}{R_{2n}}$ We know from the properties of Pascal's triangle that $R_k=2^{k-1}$

Therefore, $S_n= \frac{2^{n-1}}{2^{n}}+\frac{2^{n-2}}{2^{n+1}}+\frac{\dots}{\dots}+\frac{2^2}{2^{2n-3}}+\frac{2^1}{2^{2n-2}}+\frac{2^0}{2^{2n-1}}$ $\lim\limits_{n \to \infty}S_n=S=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\dots$ $\Rightarrow (\frac{1}{2^2})S= \frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dots$ $\Rightarrow (1-\frac{1}{2^2})S= \frac{1}{2}$ $\Rightarrow S=\lim\limits_{n \to \infty}S_n=\frac{2}{3} =\frac{a}{b}$ $\Rightarrow a+b=2+3=\boxed{5}$