Number Theory

\hspace{-16}\bf{\Rightarrow 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.........-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}}$\\\\\\ $\bf{\Rightarrow \left(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{1319}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{1318}\right)=\frac{p}{q}}$\\\\\\ $\bf{\Rightarrow \left(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{1319}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{659}\right)=\frac{p}{q}}$\\\\\\ $\bf{\Rightarrow \frac{1}{660}+\frac{1}{661}+............................+\frac{1}{1319}=\frac{p}{q}}$\\\\\\ $\bf{\Rightarrow \left(\frac{1}{660}+\frac{1}{1319}\right)+\left(\frac{1}{661}+\frac{1}{1318}\right)+.......+\left(\frac{1}{979}+\frac{1}{990}\right)=\frac{p}{q}}$\\\\\\ $\bf{\Rightarrow 1979.\left(\frac{1}{660.1319}+\frac{1}{661.1318}+......+\frac{1}{989.990}\right)=\frac{p}{q}}$\\\\\\

\hspace{-16}$Now Here $\bf{1979}$ is a prime no.\\\\\\ So $\bf{p}$ is a multiple of $\bf{1979}$\\\\\\ or $\bf{p}$ is divisible by $\bf{1979}$