Question

Find the value of the power series sum: `sum_{n=2}^∞ \frac{(-1)^n}{(2n+1)(2n+2)3^n}` ?

Answer 1

`sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2)) = sqrt3arctan(1/sqrt3) - 3/2 ln(4/3) - 17/36`

Explanation:

Consider the geometric series:

`sum_(n=0)^oo q^n = 1/(1-q)`

for `absq < 1`.

Let `q=-x^2`

`1/(1+x^2) = sum_(n=0)^oo (-1)^nx^(2n)`

Integrate term by term:

`int_0^x dt/(1+t^2) = sum_(n=0)^oo (-1)^n int_0^x t^(2n)dt`

`arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)dt`

Integrate again, considering that integrating by parts:

`int arctanx dx = xarctanx - int x/(1+x^2)dx`

`int arctanx dx = xarctanx - 1/2 int (d(1+x^2))/(1+x^2)`

`int arctanx dx = xarctanx - 1/2 ln(1+x^2)+c`

so:

`int_0^x arctant = sum_(n=0)^oo (-1)^n/(2n+1) int_0^x t^(2n+1)dt`

`xarctanx - 1/2 ln(1+x^2) = sum_(n=0)^oo (-1)^n x^(2n+2)/((2n+1)(2n+2))`

Let now `x=1/sqrt3`. As `0 < 1/sqrt3 < 1` the series is convergent in this point:

`1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = sum_(n=0)^oo (-1)^n 1/(sqrt3)^(2n+2)1/((2n+1)(2n+2))`

note that:

`(sqrt3)^(2n+2) = 3*(sqrt3)^(2n) = 3*3^n`

so:

`1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = 1/3sum_(n=0)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))`

Extract now the first two terms for `n=0` and `n=1`:

`1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = 1/6 - 1/108 +1/3sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))`

`1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) - 1/6 + 1/108 =1/3sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))`

`sqrt3arctan(1/sqrt3) - 3/2 ln(4/3) - 17/36 =sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))`