Limit of infinite sum. Let $x$ be a real number such as $0 < x < 1$, compute the following series 
$$\sum^\infty _{n=2} \frac{x^n}{n-1}$$ $$\sum^\infty _{n=2} \frac{x^n}{n+1}
$$
For instance, I made something like this : $$\frac{x^n}{n-1} = \frac{x^n}{n(1-\frac{1}{n})} \sim _{n \rightarrow \infty} \frac{x^n}{n}(1+\frac{1}{n} + o(\frac{1}{n})) = \frac{x^n}{n-1} + o(\frac{1}{n})$$
and finaly we have something like this :
$$ \sum_{n=2} ^\infty \frac{x^n}{n-1} \sim \sum^\infty _{n=2}\frac{x^n}{n} = -ln(1-x)$$
However something let me think that this result couldn't be the good one.
NB : sorry for my bad english                                                                                                                                              

Let $n^\prime = n-1$:
$\sum\limits_{n=2}^\infty = \sum\limits_{n^\prime=1}^\infty {x^{n^\prime + 1} \over n^\prime} = x \sum\limits_{n^\prime}^\infty {x^{n^\prime} \over n^\prime} = -x \ln(1-x)$.
