Convergence of $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ Would anyone happen to know what $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ converges to?  My method goes as
$$
f(x) 
= \sum_{n=1}^{\infty} \frac{x^n}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \int_0^x t^ndt
= \frac{1}{x} \int_0^x \sum_{n=1}^{\infty} t^ndt
= \frac{1}{x} \int_0^x \frac{t}{1-t}dt
= \frac{-x - \ln (1-x)}{x}
= \frac{1}{x}\ln\big(\frac{1}{1-x}\big) - 1 ~.
$$
Can anyone confirm?

$$\sum_{n=1} \frac{x^n}{n+1} = \frac{1}{x}\sum_{n=2} \frac{x^n}{n} = \frac{1}{x}\left(\sum_{n=1} \frac{x^n}{n} - x\right) = -1 - \frac{\log(1-x)}{x}$$
for $0<|x| < 1$ (the limit to $0$ exists, and equals $0$)
