I would like to prove that, for $n \in \mathbb{N}$, $$\log\bigg(1-\bigg (\frac{\overline{\theta}}{\theta}\bigg )^{n+1}\bigg) \leq \bigg | \frac{\overline{\theta}}{\theta-\overline{\theta}} \bigg |,$$
where $\theta = \frac{b+\sqrt{b^2+4c}}{2}$ and $\overline{\theta}= \frac{b-\sqrt{b^2+4c}}{2}$
The hint is to use the Taylor series for $\log(1+x)$ and bound it from above with a geometric series.
My attempt so far: Let $x= \frac{\overline{\theta}}{\theta}$ and I know that $|x| < 1$, so using Wolfram alpha I found the Taylor series for $\log(1+x)$, which is, \begin{equation}-\sum_{n=1}^{\infty} \frac{(-1)^n(-x)^n}{n} = -\sum_{n=1}^{\infty} \frac{(-1)^{2n}(x)^n}{n}=-\sum_{n=1}^{\infty} \frac{(x)^n}{n}\end{equation}
The geometric series is: $$\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}$$
Now, my question is, can I do the following? $$\bigg|-\sum_{n=1}^{\infty} \frac{x^n}{n}\bigg| \leq \bigg|\sum_{n=1}^{\infty} x^n\bigg| = \bigg |\frac{x}{1-x} \bigg |=\bigg | \frac{\overline{\theta}}{\theta-\overline{\theta}} \bigg |$$
Is it correct to do the above? If it's not correct do you have suggestions for a correct way?