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

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)) #