#f(x) = ln(1+x)-ln(1-x)#
Differentiation leads to
#fprime (x) = 1/{1+x}+1/{1-x}#
Let us consider one of the pieces : #1/{1+x}#. Differentiating this repeatedly leads to
#d/dx(1/{1+x}) = {-1}/(1+x)^2#
#d^2/dx^2(1/{1+x}) = {(-1) times (-2)}/(1+x)^3#
#d^3/dx^3(1/{1+x}) = {(-1) times (-2) times (-3)}/(1+x)^4#
The pattern is clear. We have
#d^n/dx^n(1/{1+x}) = {(-1)^n n!}/(1+x)^{n+1}#
#d^n/dx^n(1/{1-x}) = {n!}/(1-x)^{n+1}#
and similarly
This means that
#f^(n+1)(x) =d^n/dx^n(1/{1+x})+d^n/dx^n(1/{1-x}) #
# = n![{(-1)^n }/(1+x)^{n+1}+1/(1+x)^{n+1}]#
To find the Maclurin polynomial, we need to evaluate #f^(n)(0)#. This simply is
#f^(n)(0) = (n-1)![1+(-1)^{n-1}]#
So, #f^(n)(0)# is #2(n-1)!# for odd #n# and it vanishes for even #n#, and the coefficients of the Maclurin polynomial are #{f^(n)(0)} /{n!} = 2/n# for odd #n# and 0 for even #n#
Thus the Maclurin polynomial is
#P_n(x) = 2x+2/3x^3+2/5x^5+... = 2 sum_{i=0}^{[n/2]-1}x^{2i+1}/{2i+1}#
An alternative approach would be to start from the GP
#1+x+x^2+...+x^n = {1-x^{n+1}}/{1-x}#
to get
#1-x+x^2+...+(-1)^nx^n = {1-(-1)^{n+1}x^{n+1}}/{1+x}#
so that
#2 + 2x^2 + ...+[1+(-1)^n]x^n = 1/{1+x}+1/{1-x}-x^{n+1}[1/{1-x}+(-1)^{n+1}/{1+x}]#
Integrating both sides gives the Maclurin series for #f(x)# (the constant of integration being found from #f(0)=0#. This approach also shows that the remainder term is
#R_n(x) = int_0^x y^{n+1}[1/{1-y}+(-1)^{n+1}/{1+y}]dy#
This form may be helpful in proving that the remainder vanishes as #n to infty# for #|x|<1#
To find #ln(2)# using this series, use #x= 1/3#
#f(1/3) = ln({1+1/3}/{1-1/3})= ln(2) = 2[1/{1cdot3}+1/{3cdot 3^3}+1/{5cdot 3^5}+1/{7cdot3^7}...] ~~ 0.6931#
