## anonymous 5 years ago what is the sum of 1/(k^2)-1 with the top limit as infinity and bottom limit is k=2 why is the answer 3/4?

Notice that $\frac{1}{k^2-1}-\frac{1/2}{k-1}-\frac{1/2}{k+1}$ Then consider the partial sum $$S_n$$ (it is a telescoping sum). Then compute the limit.