Question

How do you find the Maclaurin series of `f(x)=ln(1+x)` ?

Answer 1

The Maclaurin series of `f(x)=ln(1+x)` is:
`f(x)=sum_{n=0}^{infty}(-1)^{n}{x^{n+1}}/{n+1}`,
where `|x|<1`.

First, let us find the Maclaurin series for
`f'(x)=1/{1+x}=1/{1-(-x)}`.

Remember that
`1/{1-x}=sum_{n=0}^{infty}x^n` if `|x|<1`.
(Note: This can be justified by viewing it as a geometric series.)

By replacing `x` by `-x`,
`f'(x)=1/{1-(-x)}=sum_{n=0}^{infty}(-x)^n=sum_{n=0}^{infty}(-1)^nx^n`

By integrating using Power Rule,
`f(x)=intsum_{n=0}^{infty}(-1)^nx^n dx=sum_{n=0}^{infty}(-1)^n{x^{n+1}}/{n+1}+C`
(Note: integration can be done term by term.)

Since `f(0)=ln[1+(0)]=0`,
`f(0)=sum_{n=0}^{infty}(-1)^{n}{(0)^{n+1}}/{n+1}+C=C=0`.

Hence,
`f(x)=sum_{n=0}^{infty}(-1)^{n}{x^{n+1}}/{n+1}`.