直接分項對消就可以了 \(\displaystyle \frac{1}{k^2-n^2} = \frac{1}{2n}\left(\frac{1}{k-n}-\frac{1}{k+n}\right)\)
所求對消之後剩下的 \(\displaystyle =\frac{1}{2n}\left(\frac{1}{1-n}+\frac{1}{2-n}+\cdots+\frac{1}{-1}+\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{2n}\right)\)
\(\displaystyle =\frac{1}{2n}\left(\frac{1}{n}+\frac{1}{2n}\right)=\frac{3}{4n^2}\)
