#6/( n(n-3) ) = A/ n + B/(n-3) #
#=> 6 = A(n-3) + Bn #
# n = 3 #
#=> 6 = 3B => B = 2 #
# n = 0 #
# 6 = -3A => A = -2 #
#=> sum_(n=5) ^(oo) (2/(n-3) -2/n) #
Using method of differences:
# n = 5 : 2/2 cancel(- 2/5 )#
#n = 6: 2/3 cancel(- 2/6) #
#n = 7: 2/4 cancel(- 2/7) #
#n = 8: cancel( 2/5) cancel(- 2/8) #
#.#
#.#
#.#
#.#
#n = r-3 : cancel(2/(r-6)) cancel(- 2/(r-3) #
#n = r-2 :cancel( 2/(r-5)) - 2/(r-2)#
#n = r-1 : cancel(2/(r-4)) - 2/(r-1) #
#n = r: cancel(2/(r-3)) - 2/r #
#=> sum_(n =5 ) ^r 6/(n^2 - 3n ) #
#= 2/2 + 2/3 + 2/4 - 2/(r-2) - 2/(r-1) - 2/r #
#=> sum_(n =5 ) ^oo 6/(n^2 - 3n ) = lim_(r to oo) sum_(n =5 ) ^r 6/(n^2 - 3n )#
# = 2/2 +2/3 + 2/4 #
# = 13/6 #