How to solve sum_(n=1)^oo 1/((3n-2)(3n+1))n=11(3n2)(3n+1) using partial fractions?

sum_(n=1)^oo 1/((3n-2)(3n+1))n=11(3n2)(3n+1)
There is an infinity sign on the top of the summation sign and n=1 on the bottom.
Please write in details; I don't get how the answer is 1/12112.

1 Answer
Nov 25, 2017

sum_(n=1)^oo 1/((3n-2)(3n+1)) = 1/3n=11(3n2)(3n+1)=13

Explanation:

1/((3n-2)(3n+1)) = A/(3n-2)+B/(3n+1)1(3n2)(3n+1)=A3n2+B3n+1

So:

1 = A(3n+1)+B(3n-2)1=A(3n+1)+B(3n2)

Putting n=2/3n=23 this becomes:

1 = 3A" "1=3A so " "A=1/3 A=13

Putting n=-1/3n=13 it becomes:

1 = -3B" "1=3B so " "B=-1/3 B=13

Then:

sum_(n=1)^N 1/((3n-2)(3n+1)) = sum_(n=1)^N (1/(3(3n-2))-1/(3(3n+1)))Nn=11(3n2)(3n+1)=Nn=1(13(3n2)13(3n+1))

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = sum_(n=1)^N 1/(3(3n-2)) - sum_(n=1)^N 1/(3(3n+1))Nn=11(3n2)(3n+1)=Nn=113(3n2)Nn=113(3n+1)

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = sum_(n=1)^N 1/(3(3n-2)) - sum_(n=2)^(N+1) 1/(3(3n-2))Nn=11(3n2)(3n+1)=Nn=113(3n2)N+1n=213(3n2)

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - 1/(3(3N+1)

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3 - 1/(3(3N+1)

So:

sum_(n=1)^oo 1/((3n-2)(3n+1)) = lim_(N->oo) sum_(n=1)^N 1/((3n-2)(3n+1))

color(white)(sum_(n=1)^oo 1/((3n-2)(3n+1))) = lim_(N->oo) (1/3 - 1/(3(3N+1))) = 1/3

Note that:

sum_(n=color(blue)(2))^oo 1/((3n-2)(3n+1)) = 1/(3(3(color(blue)(2))-2)) = 1/12