An Easy One

$\sum_{x=1}^{\infty} \frac{3}{6x^{2}+6x} = \frac{1}{2} \sum_{x=1}^{\infty} \frac{1}{x^{2}+x} = \frac{1}{2} \sum_{x=1}^{\infty} \frac{1}{x(x+1)}$ Using partial fraction decomposition : $=\frac{1}{2} \sum_{x=1}^{\infty} \frac{1}{x}- \frac{1}{(x+1)}$ The series $\sum_{x=1}^{\infty} \frac{1}{x}- \frac{1}{(x+1)}$ converges (telescopes) to $1$ . Multiplied by $\frac{1}{2}$ gives us $\frac{1}{2}$ Rounding to the nearest ones gives us $\boxed{1}$ .