Find the sum of the series: #sum_{n=5}^ ∞ 6 / (n^2 - 3n)# ?

#sum_{n=5}^ ∞ 6 / (n^2 - 3n)#
Also, I was wondering if it can be solved as a telescoping series?

1 Answer
Rhys
Jun 1, 2018

# = 13/6 #

Explanation:

#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 #

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