Question

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

`sum_(n=1)^oo 1/((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 `1/12`.

Answer 1

`sum_(n=1)^oo 1/((3n-2)(3n+1)) = 1/3`

Explanation:

`1/((3n-2)(3n+1)) = A/(3n-2)+B/(3n+1)`

So:

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

Putting `n=2/3` this becomes:

`1 = 3A" "` so `" "A=1/3`

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

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

Then:

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

`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))`

`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`