# Math Help - Maclaurin series - check my work

1. ## Maclaurin series - check my work

Cn someone check my Maclaurin series for:

ln(1+x)

I got:

$\sum^{\infty}_{n=0} (-1)^n \frac{(n-1)!}{(1+x)^n}$

2. Originally Posted by mollymcf2009
Cn someone check my Maclaurin series for:

ln(1+x)

I got:

$\sum^{\infty}_{n=0} (-1)^n \frac{(n-1)!}{(1+x)^n}$
$f(x)=\ln(1+x)$

Then

$f'(x)=\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty}(-1)^nx^n$

Now integrating both sides we get

$f(x)=C+\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}$

And since $f(0)=\ln(1)=0=C$ we get

$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}$