How to solve sum_(n=1)^oo 1/((3n-2)(3n+1))∞∑n=11(3n−2)(3n+1) using partial fractions?
sum_(n=1)^oo 1/((3n-2)(3n+1))∞∑n=11(3n−2)(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 .
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 Answer
Explanation:
1/((3n-2)(3n+1)) = A/(3n-2)+B/(3n+1)1(3n−2)(3n+1)=A3n−2+B3n+1
So:
1 = A(3n+1)+B(3n-2)1=A(3n+1)+B(3n−2)
Putting
1 = 3A" "1=3A so" "A=1/3 A=13
Putting
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)))N∑n=11(3n−2)(3n+1)=N∑n=1(13(3n−2)−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))N∑n=11(3n−2)(3n+1)=N∑n=113(3n−2)−N∑n=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))N∑n=11(3n−2)(3n+1)=N∑n=113(3n−2)−N+1∑n=213(3n−2)
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