Find a power series representation for $f(x)=\ln(x+1)$.

Question

I need some help with checking this exercise. Thanks.

Find a power series representation for $f(x)=\ln(x+1)$.

1. Find $f'(x)$

$$f'(x)=\frac{1}{x+1}$$ $$f'(x)=\frac{1}{1+x}$$ $$f'(x)=\frac{1}{1-(-x)}$$

Taking in consideration that $\frac{a_1}{1-r}=\Sigma_{n=0}^\infty a_1 r^n$, then $a_1=1$ and $r=-x$

2. $f'(x)$ as a series

Now we have:

$$f'(x)=\frac{1}{1-(-x)}$$ $$f'(x)=\Sigma_{n=0}^\infty [1\cdot (-x)^n]$$ $$f'(x)=\Sigma_{n=0}^\infty [(-1\cdot x)^n]$$ $$f'(x)=\Sigma_{n=0}^\infty [(-1)^n\cdot x^n]$$

3. $f(x)$ as a series

$$f(x)=\int f'(x) dx$$ $$f(x)=\int\Sigma_{n=0}^\infty [(-1)^n\cdot x^n]dx$$ $$f(x)=\Sigma_{n=0}^\infty [(-1)^n\cdot \frac{x^{n+1}}{n+1}]+C$$

Since, $f(0)=\ln(0+1)=\ln(1)=0$, then $C=0$. Therefore,

$$f(x)=\Sigma_{n=0}^\infty [(-1)^n\cdot \frac{x^{n+1}}{n+1}]+0$$

$$f(x)=\Sigma_{n=0}^\infty [(-1)^n\cdot \frac{x^{n+1}}{n+1}]$$

My question is, do I leave it like this last previous step, or do I need to re-write it as

$$f(x)=\Sigma_{n=1}^\infty [(-1)^{n-1}\cdot \frac{x^{n+1-1}}{n+1-1}]$$

$$f(x)=\Sigma_{n=1}^\infty [(-1)^{n-1}\cdot \frac{x^{n}}{n}]$$

Or can I write it either way with $n=0$ or $n=1$?