How to find generating function of a sequence

Question

Need help to find generating function of the following series.

$$\sum_{n=0}^{\infty} \frac{x^n}{n+1}$$

Thanks!

Answer 1

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

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

$\frac{d}{dx}xG(x) = \sum_{n=0}^{\infty} \frac{1}{n+1}\frac{d}{dx}x^{n+1}$

$\frac{d}{dx}xG(x) = \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$

This implies $G(x) = -\frac{\log(1-x)}{x}$

Answer 2

$$ \int_0^x t^{n} dt = \frac{x^{n+1}}{n+1} $$

we then have $$ \sum_{n=0}^\infty \frac{x^n}{n+1} = \frac{1}{x}\sum_{n=0}^\infty\int_0^x t^n dt = \frac{1}{x}\int_0^x dt \sum_{n=0}^\infty t^n $$ The last sum is the geometric series which has a known closed form for $|t| < 1$

Answer 3

$$=\frac{1}{x}\sum \frac{x^{n+1}}{n+1}=-\frac{1}{x}\log(1-x)$$