Binomial Coefficient Challenge 5!

Note that $\begin{aligned} \frac{S(m)}{2m+1} & = \sum_{k=0}^m \binom{m}{k}\frac{(-2)^k}{2m+1-k} \; = \; \int_0^1 \sum_{k=0}^m \binom{m}{k} (-2)^k t^{2m-k}\,dt \\ & = \int_0^1t^{2m}(1 - 2t^{-1})^m\,dt \; =\; \int_0^1(t^2-2t)^m\,dt \; = \; \int_0^1 (u^2-1)^m\,du \end{aligned}$ for any $m \ge 0$ . Thus $\begin{aligned} \sum_{m=0}^\infty \frac{S(m)}{2m+1}x^{2m+1} & = \sum_{m=0}^\infty x^{2m+1}\int_0^1 (u^2-1)^m\,du \; = \; \int_0^1 \frac{x\,du}{1 - x^2(u^2-1)} \\ & = \int_0^1 \frac{x\,du}{1+x^2 - x^2u^2} \; = \; \frac{1}{\sqrt{1+x^2}} \tanh^{-1}\frac{x}{\sqrt{1+x^2}} \end{aligned}$ for any $|x| < 1$ . Differentiating within the radius of convergence, we obtain $\sum_{m=0}^\infty S(m)x^{2m} \; = \; \frac{1}{1+x^2}\Big[ 1 - \frac{x}{\sqrt{1+x^2}}\tanh^{-1}\frac{x}{\sqrt{1+x^2}}\Big] \hspace{1cm} |x| < 1$ Putting $x = \tfrac{1}{\sqrt{2}}$ gives $\sum_{m=0}^\infty S(m) 2^{-m} \; = \; \tfrac23\Big[1 - \tfrac{1}{\sqrt{3}}\tanh^{-1}\tfrac{1}{\sqrt{3}}\Big]$ making the answer $2 + 3 + 3 = \boxed{8}$ .

From the integral representation of $S(m)$ it is easy to show that $S(m) \; = \; \frac{(-4)^m}{\binom{2m}{m}} \hspace{2cm} m \ge 0$