If `S=sum_(n=2)^(oo) (3n^2+1)/((n^2-1)^3)` then 9/4Sequals

A

a,b,c

B

a,c ,b

C

b,c,a

D

b,a,c

Verified by Experts

The correct Answer is:

D

Similar Questions

Explore conceptually related problems

sum_(n=1)^(oo) (2n)/(n!)=

sum_(n=1) ^(oo) (1)/((2n-1)!)=

sum_(n=1)^(oo)(2 n^2+n+1)/((n!)) =

sum_(n=1)^(oo)(1)/(2n-1)*x^(2n)=

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

sum_(n=1)^(oo) ((log_ex)^(2n-1))/((2n-1)!)=

sum_(n=1)^(oo) (1)/(2n(2n+1)) is equal to

p=sum_(n=0)^(oo) (x^(3n))/((3n)!) , q=sum_(n=1)^(oo) (x^(3n-2))/((3n-2)!), r = sum_(n=1)^(oo) (x^(3n-1))/((3n-1)!) then p + q + r =

The sum sum_(n=1)^oo tan^(-1)(3/(n^2+n-1)) is equal to

The value of sum_(n=1)^oo\ (n^2+1)/((n+2)n!) is