Question

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

Answer 1

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)`