To find a Maclaurin series for #ln((1+x)/(1-x))# from scratch, we first need to take note of expressing a function as an infinite sum centered at #x= 0#.
In order to do this, we write
#f(x) = #
#f(0) + f^(1)(0)/(1!)x + f^(2)(0)/(2!)x^2 + f^(3)(0)/(3!)x^3 + ... = sum_(n=0)^(∞) f^(n)(0) x^(n) / (n!)#
This infinite sum suggests that we'd have to calculate some derivatives, which I will do for three terms - hopefully enough to spot a pattern.
By the properties of logarithms, we have
#f(x) = ln((1+x)/(1-x)) = ln(1+x) - ln(1-x)#
Derivative calculations:
#f^(1)(x) = 1/(1+x) - (-1)/(1-x) = 1/(1+x) + 1/(1-x)#
# = (1cancel(-x)+1cancel(+x))/(1cancel(-x+x)-x^2) = 2/(1-x^2) = 2(1-x^2)^(-1)#
#f^(2)(x) =-2(1-x^2)^(-2)(-2x) = 4x(1-x^2)^(-2) = (4x)/(1-x^2)^(2)#
#f^(3)(x) =4[1(1-x^2)^(-2)+x*-2(1-x^2)^(-3)(-2x)]#
#= 4/(1-x^2)^(2) + (16x^2)/(1-x^2)^(3)#
Substituting #0# into our derivatives we get
#f(0) = 0#
#f^(1)(0) = 2#
#f^(2)(0) = 0#
#f^(3)(0) = 4#
#...#
#f^(4)(0) = 0#
#f^(5)(0) = 48#
Actually, it turns out that only odd derivatives at #x=0# give an actual value, and so to make up for a missing term, I've used a computing device to calculate the fourth and fifth derivative, as shown above.
Using our #f^(n)(0)#-values to construct a Maclaurin series, we write
#ln((1+x)/(1-x)) = 0 + 2/(1!) x + 0/(2!) x^2 + 4/(3!) x^3 + 0/(4!) x^4 + 48/(5!) x^5 + ...#
As we can see, a few terms just cancel out, leaving us with
#ln((1+x)/(1-x)) = 2/(1!) x + 4/(3!) x^3 + 48/(5!) x^5 + ...#
Simplifying the denominators on each term (the factorials, that is), we end up with
#ln((1+x)/(1-x)) = 2x + 4/(6) x^3 + 48/(120) x^5 + ...#
We can actually simplify a few fractions and factor out a #2# out of each term, leaving us with
#ln((1+x)/(1-x)) = 2(x + 1/(3) x^3 + 1/(5) x^5 + ...)#
As we may notice, we can see that the power and the denominator on each term is increasing by #2#, so we end up with
#ln((1+x)/(1-x)) = 2(x + 1/(3) x^3 + 1/(5) x^5 + ...) = 2sum_(n=0)^(∞) x^(2n+1) / (2n+1)#
