Find `sum_(n=1)^(oo)1/(n^2+5n+6)`

If theta in (pi//4, pi//2) and sum_(n=1)^(oo)(1)/(tan^(n)theta)=sin theta + cos theta , then the value of tan theta is

The expression sum_(n=1)^(oo)cot^(-1)(n^(2)-3n+3) simplifies to

Find the sum Sigma_(r=1)^(oo)(3n^2+1)/((n^2-1)^3)

Find the sum_(k=1)^(oo) sum_(n=1)^(oo)k/(2^(n+k)) .

The value of sum sum_(n=1)^(oo)cot^(-1)(((n^(2)+2n)(n^(2)+2n+1)+1)/(2n+2)) is equal to

Find sum_(k=1)^(n)(1)/(k(k+1)) .