Question

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?

Answer 1

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.