, 2010

Solution (by Titu Andreescu). We have
$$\begin{aligned} \frac{2}{k(k+1)(k+2)} &= \frac{(k+2)-k}{k(k+1)(k+2)} \\ &= \frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)} \\ &= \left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \\ &= \frac{1}{k} + \frac{1}{k+1} + \frac{1}{k+2} - \frac{3}{k+1}. \end{aligned}$$
Hence
$$\begin{aligned} 2S_q &= \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{q} + \frac{1}{q+1} + \frac{1}{q+2}\right) - 3\left(\frac{1}{3} + \frac{1}{6} + \dots + \frac{1}{q+1}\right) \\ &= \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{\frac{3p-1}{2}}\right) - \left(1 + \frac{1}{2} + \dots + \frac{1}{\frac{p-1}{2}}\right), \end{aligned}$$
and so
$$\begin{aligned} 1 - \frac{m}{n} &= 1 + 2S_q - \frac{1}{p} \\ &= \frac{1}{\frac{p+1}{2}} + \dots + \frac{1}{p-1} + \frac{1}{p+1} + \dots + \frac{1}{\frac{3p-1}{2}} \\ &= \left( \frac{1}{\frac{p+1}{2}} + \frac{1}{\frac{3p-1}{2}} \right) + \dots + \left( \frac{1}{p-1} + \frac{1}{p+1} \right) \\ &= \frac{p}{\left( \frac{p+1}{2} \right) \left( \frac{3p-1}{2} \right)} + \dots + \frac{p}{(p-1)(p+1)}. \end{aligned}$$
Because the denominator of each term in the sum is relatively prime to $p$, it follows that $n-m$ is divisible by $p$, as desired.