Question

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

Answer 1

`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.