Substitute for \(\displaystyle 2k=t \; \) , the sum becomes \(\displaystyle s=\sum_{t=2}^{60} \dfrac{2}{t^2 -1 } \).

Solve the equation for \(\displaystyle p \: \) : \(\displaystyle \dfrac{2}{t^2 -1 }=p[\dfrac{1}{t-1}-\dfrac{1}{t+1} ] \; \Rightarrow \) \(\displaystyle p=1\).

Now the expression inside the sum(general term) becomes \(\displaystyle \dfrac{1}{t-1}-\dfrac{1}{t+1}\).

\(\displaystyle s =\displaystyle \sum_{t=2}^{60} [\dfrac{1}{t-1}-\dfrac{1}{t+1}]=\sum_{t=2}^{60}\dfrac{1}{t-1} -\sum_{t=2}^{60}\dfrac{1}{t+1}\).

\(\displaystyle s=[1+1/2+1/3+\dotsc +1/59]-[1/3+1/4+1/5+\dotsc +1/59 +1/60+1/61]=1+1/2-1/60-1/61\).