It is simpler than it looks!

$S=\sum_{n=1}^\infty \dfrac{B(n,n)}{2n}=\sum_{n=1}^\infty \int_0^1 \dfrac{(x(1-x))^{n-1}}{2n} dx\\=\int_0^1\sum_{n=1}^\infty \dfrac{(x(1-x))^{n-1}}{2n} dx=\dfrac{1}{2}\int_0^1 \ln(1-x(1-x))(x(1-x))^{-1}$ Using partial fractions $S=\dfrac{1}{2}\left(\int_0^1 \dfrac{\ln(x^2-x+1)}{x}dx+\int_0^1 \dfrac{\ln(x^2-x+1)}{1-x}dx\right)$ In the second integral use $\int_b^a f(x) dx=\int_b^a f(a+b-x)dx$ to get both integrals are the same. We have $S=\int_0^1 \dfrac{\ln(x^2-x+1)}{x}dx=\int_0^1 \dfrac{\ln(x+\omega)}{x}+\dfrac{\ln(x+\omega^2)}{x}dx$ Lets try to finding the indefinite integral first $\int \dfrac{\ln(x+\omega)}{x}+\dfrac{\ln(x+\omega^2)}{x}dx=\int \dfrac{\ln(\omega)+\ln(1+x\omega^2)}{x}+\dfrac{\ln(\omega^2)+\ln(1+x\omega)}{x}dx\\=\int \dfrac{\ln(1+x\omega^2)}{x\omega^2} \omega^2dx+\int \dfrac{\ln(1+x\omega)}{x\omega}\omega dx=Li_2(-x\omega^2)+Li_2(-x\omega)\\=\sum_{n=1}^\infty \dfrac{(-x)^n(\omega^n+\omega^{2n})}{n^2}$ SInce $\omega^n+\omega^{2n}=\begin{cases} 2, 3|n\\ -1, otherwise\end{cases}$ so we can wirte this as $\sum_{n=1}^\infty \dfrac{-(-x)^n}{n^2}+3\sum_{n=1}^\infty \dfrac{(-x)^{3n}}{(3n)^2}=-Li_2(-x)+\dfrac{1}{3}Li_2(-x^3)$ Note, $Li_s(-1)=(2^{1-s}-1)Li_s(1)=\zeta(s)(2^{1-s}-1)$ $S=-Li_2(-1)+\dfrac{1}{3}Li_2(-1)=\dfrac{1}{2}\zeta(2)-\dfrac{1}{6}\zeta(2)=\dfrac{1}{3}\zeta(2)$ $1+3+2=\boxed{6}$

The Taylor Series of $\arcsin ^{ 2 }{ x }$ about 0 is: $\arcsin ^{ 2 }{ x } =\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }\left( \begin{matrix} 2n \\ n \end{matrix} \right) } } { \left( 2x \right) }^{ n }$

Substitute: $x=\frac{1}{2}$ .

Therefore, $\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }\left( \begin{matrix} 2n \\ n \end{matrix} \right) } } = \frac{\pi^2}{18}=\frac { 1 }{ 3 } \zeta \left( 2 \right)$ .

Here, $A=1, \ B=3, \ C=2$ .

$\boxed{A+B+C=1+3+2=6}$