A Difficult One.

The $n$ th term in the series is $\frac{(2n)!}{2^{2n+1}\,n!\,(n+1)!} \; = \; \frac{1}{2^{2n}}\binom{2n}{n} - \frac{1}{2^{2n+2}}\binom{2n+2}{n+1}$ and hence the sum to $N$ terms is $S_N \; = \; \frac12 - \frac{1}{2^{2N+2}}\binom{2N+2}{N+1}$ Thus $S_{10} \; = \; \frac12 - \frac{1}{2^{22}}\binom{22}{11} \; = \; \frac{173965}{524288}$ making the answer $173965 + 524288 + 271716 = \boxed{969969}$ .

Since $2^{-2n}\binom{2n}{n} \sim \frac{1}{\sqrt{n\pi}}$ as $n \to \infty$ , using Stirling's approximation, we deduce that $\lim_{N \to \infty}S_N \; = \; \tfrac12$ .

$S_{10} = \frac{1}{2.4} + \frac{1.3}{2.4.6} + \cdots + \frac{1.3.6.\cdots. 21}{2.4.6.\cdots. 22} \\ S_{10} = \frac{4-3}{2.4} + \frac{1.3(6-5)}{2.4.6.} +\frac{1.3.5.(8-9)}{2.4.6.8.10}+ \cdots + \frac{1.3.6.\cdots. 21.(22-21)}{2.4.6.\cdots. 22} \\ S_{10} = \frac{4}{2.4} -\frac{3}{2.4} + \frac{3.6}{2.4.6} + \frac{3.5.8}{2.4.6.8} -\frac{3.5.7}{2.4.6.8}+\cdots - \dfrac{1.3.5.\cdots21 }{2.4.6\cdots 22}\\ \implies S_{10} = \dfrac{1}{2} - \dfrac{(21)!!}{(22)!!} = \frac{1}{2} - \frac{21!}{4.(11!)^2} =\frac{1}{2} - \frac{88179}{524288} \\ \therefore S_{10} = \frac{173965}{524288} = \frac{a}{b} \implies S_{\infty} = \frac{1}{2}$ Hence, the sum of $a+b +271716 = 173965 + 524288+271716= 969969$ .