I am currently reading this paper by Hastings, which proves that the entanglement entropy of certain one dimensional systems follows an area law. Let $\xi_0 = \xi' 2 \ln[2C_1(\xi)]$. Since $C_1(\xi)$ is an input of the natural logarithm function, we see that $C_1(\xi) > 0$. The goal is to show that $$ \sum_{n = 0}^\infty 2C_1(\xi) e^{-2^n\xi_0/\xi'}\frac{\xi_0}{\xi'} \le \sum_{n = 0}^\infty e^{-2n} \le 2 $$
Defining $x \equiv2C_1(\xi)$, I was able to show that $$ \sum_{n = 0}^\infty 2C_1(\xi) e^{-2^n\xi_0/\xi'}\frac{\xi_0}{\xi'} = \sum_{n = 0}^\infty \frac{2\ln x}{x^{2^{n + 1} - 1}} $$
Plugging in specific values for $n$ allowed me to see why $\frac{2\ln x}{x^{2^{n + 1} - 1}}$ could be bounded above by $e^{-2n}$, but I could not prove it rigorously.
How does one show that for all nonnegative integers n, $\frac{2\ln x}{x^{2^{n + 1} - 1}} \le e^{-2n}$?
This is another way to show that the entire series to the left is bounded above by 2. I basically skipped the middle sum. Here is a sketch.
For each $n$ $$ \frac{2\ln x}{x^{2^{n + 1} - 1}} \le \frac{2\ln x}{x^{2^{n + 1} - 1}}\bigg|_\text{max} \le \frac{1}{2^n} $$
The maximum value can be obtained by differentiating $\frac{2\ln x}{x^{2^{n + 1} - 1}}$ with respect to $x$, setting it to zero, solving for $x$ and plugging the special value of $x$ into $\frac{2\ln x}{x^{2^{n + 1} - 1}}$. One can verify that this x gives rise to a maximum