Find the sum of the serie

Question

The serie  2/[(n+2)(n-1)]  (n starting from 3 until infinite)

Answer 1

Hi Edona,

sum from 3 to ∞

∑_n=3 ^∞  2/[(n+2)(n-1)]

Use partial fraction methods to split the fraction.

2/[(n+2)(n-1)] = A/(n +2)  + B/(n-1) 

                    2 = A(n-1) + B (n+2)

If n = 3,       2 = 2A + 5B

If n =4,        2 = 3A + 6B

A= -2/3, B =2/3

2/[(n+2)(n-1)] = (-2/3)/(n+2) + (2/3)/(n-1)

                       =(2/3)[1/(n-1) - 1/(n +2)]

Expand the first few  

 

∑2/[(n+2)(n-1)] = (2/3){[1/2 -1/5] + [1/3-1/6]+[1/4-1/7]+[1/5-1/8] +[1/6-1/9]+[1/7-1/10]+[1/8-1/11]+[1/9-1/12]+[1/10-1/13] +[1/11-1/14] + [1/12-1/15]+[1/13-1/16]+[1/14-1/17]+ ...

                       = (2/3)[1/2+1/3+1/4 + 0 + 0 + 0 + ...]

                       =(2/3)[13/12]

∑2/[(n+2)(n-1)] = 13/18