What is the power series representation of #ln((1+x)/(1-x))#?

1 Answer
Dec 4, 2016

# ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)#

Explanation:

I would use the following

  1. The log rule; #log(A/B) = logA-logB#
  2. The known power series :
    #ln(1+x) = 1-x^2/2+x^3/3-x^4 ...=sum_(n=1)^oo(-1)^(n+1)x^n/n#

So:
# \ \ \ \ \ ln((1+x)/(1-x)) =ln(1+x)-ln(1-x) #
# :. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-(-x)^2/2+(-x)^3/3-(-x)^4 + ...} #
# :. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-x^2/2-x^3/3-x^4 + ...} #
# :. ln((1+x)/(1-x)) =1-x^2/2+x^3/3-x^4 + ... - 1+x^2/2+x^3/3+x^4 + ... #
# :. ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)#