Question

series

Answer 1

1/(3k-1)(3k+2) = A/(3k-1)+B/(3k+2)
1=(3k+2)A+(3k-1)B
put k=-2/3, -3B=1, B=-1/3
put k=1/3, 3A=1, A=1/3
1/(3k-1)(3k+2) = 1/3(3k-1)-1/3(3k+2)
\[\sum_{1}^{n}1/3(3k-1)-1/3(3k+2)\]
=1/3(2)-1/3(3n+2)

take limit, then
ans =1/6