How do you derive the maclaurin series for #f(x)=ln(1+x)#?

1 Answer
Apr 17, 2017

#sum_(n=1)^(infty)(-1)^(n-1)x^n/n#

Explanation:

#f(x)=ln(1+x)#

By differentiating w.r.t. #x#,

#f'(x)=1/(1+x)=1/(1-(-x))#

By viewing it as the sum of a geometric series,

#f'(x)=sum_(n=0)^(infty)1cdot(-x)^n=sum_(n=0)^(infty)(-1)^nx^n#

By integrating w.r.t #x# term by term,

#f(x)=sum_(n=0)^(infty)(-1)^n(x^(n+1))/(n+1)+C#

Since #f(0)=0# gives #C=0#,

#f(x)=sum_(n=0)^(infty)(-1)^n(x^(n+1))/(n+1)#

By shifting the indices by #1#,

#f(x)=sum_(n=1)^(infty)(-1)^(n-1)x^n/n#

I hope that this was clear.