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

1 Answer
Jun 7, 2018

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))