How do you find the taylor series for ln(1+x)?

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Answer 1 · 2017-03-11T04:47:42

Start with the basic geometric series:

#1/(1-x)=sum_(n=0)^oox^n#

Replacing #x# with #-x#:

#1/(1+x)=sum_(n=0)^oo(-x)^n=sum_(n=0)^oo(-1)^nx^n#

Note that integrating #1/(1+x)# gives #ln(1+x)+C#:

#int_0^x1/(1+t)dt=sum_(n=0)^oo(-1)^nint_0^xt^ndt#

#ln(1+x)=C+sum_(n=0)^oo(-1)^nx^(n+1)/(n+1)#

Letting #x=0# shows that #C=0#:

#ln(1+x)=sum_(n=0)^oo(-1)^nx^(n+1)/(n+1)#