Summation of series $\sum_{x=1}^{\infty}\frac{-\theta^x}{\ln(1-\theta)}$

Question

How should I simplify $$\sum_{x=1}^{\infty}\frac{-\theta^x}{\ln(1-\theta)}$$ into $$\frac{-\theta}{(1-\theta)\ln(1-\theta)}$$

I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.

Answer 1

Well

$$\sum_{x=1}^{\infty}\frac{-\theta^x}{\ln(1-\theta)}=\frac{-1}{\ln(1-\theta)}\sum_{x=1}^{\infty}{\theta^x}=\frac{-\theta}{\ln(1-\theta)}\sum_{x=0}^{\infty}{\theta^x}=\frac{-\theta}{\ln(1-\theta)}\cdot\frac{1}{1-\theta}$$

Answer 2

Just because of the formula $$\sum_{n=k}^{\infty}u^n=u^k\sum_{n=0}^{\infty}u^n=\frac{u^k}{1-u}.$$