\(a_1=\dfrac{2}{1\cdot 2 \cdot 3}=\dfrac{1}{3}\\ a_1+a_2 = \dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}=\dfrac{5}{12}\\ a_1+a_2+a_3=\dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}+\dfrac{2}{3\cdot 4\cdot 5}=\dfrac{9}{20}\\ a_1 + a_2 + a_3 + a_4 = \dfrac{2}{1\cdot 2\cdot 3} + \dfrac{2}{2\cdot 3\cdot 4}+\dfrac{2}{3\cdot 4\cdot 5}+ \dfrac{2}{4\cdot 5\cdot 6}=\dfrac{7}{15}\\ a_1 + a_2 + ... + a_5 = \dfrac{7}{15}+\dfrac{2}{5\cdot 6 \cdot 7}=\dfrac{10}{21}\\ a_1 + a_2 + ... + a_6 = \dfrac{10}{21} + \dfrac{2}{6\cdot 7 \cdot 8}=\dfrac{27}{56}\)
It seems like it will not exceed 1/2.
So the answer is 1/2