# $\ln(1-x^n)/\ln(1-x)$

**Claim** The function $f_n(x) = \frac{\ln(1-x^n)}{\ln(1-x)}$ is non-decreasing on the interval $x \in (0,1)$ for any $n \in \mathbb{N}$.

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*Proof*. This claim is trivial for $n = 1$, as $f_n(x) = 1$. For $n \geq 2$, we use the fact that $(1-x^n) = (1-x)\cdot \sum_{i=0}^{n-1} x^i$ to expand:

$$f_n(x) = \frac{\ln(1-x) + \ln(\sum_{i=0}^n x^i)}{\ln(1-x)} = 1 + \frac{\ln(\sum_{i=0}^n x^i)}{\ln(1-x)}$$

$f_n(x)$ is non-decreasing on $(0,1)$ iff $g_n(x) = \frac{\ln(\sum_{i=0}^n x^i)}{\ln(1-x)}$ is non-decreasing on $(0,1)$. Applying the derivative test:

$$\ln^2(1-x) \frac{d}{dx} g_n(x) = \frac{\ln(\sum_{i=0}^{n-1}x^i)}{1-x} + \frac{\ln(1-x)}{\sum_{i=0}^{n-1}x^i} \cdot \frac{1-x^n -nx^{n-1}(1-x)}{(1-x)(1+x)}$$

Since $\ln(1-x) < 0$ on $x \in (0,1)$, we see that $\frac{d}{dx} g_n(x) \geq 0$ on $x \in (0,1)$ iff:

$$\ln(\sum_{i=0}^{n-1}x^i) \geq \frac{|\ln(1-x)|}{\sum_{i=0}^{n-1}x^i} \cdot \frac{1-x^n -nx^{n-1}(1-x)}{1 +x}$$

Setting $y = \sum_{i=1}^{n-1} x^i$, this inequality (i.e., what we need to show) is equivalent to:

$$\frac{(1+y)\ln(1+y)}{|\ln(1-x)|} \geq \frac{1-x^n -nx^{n-1}(1-x)}{1 +x} $$

The fact that $\ln(1+z) \geq \frac{z}{1+z}$ for $z > -1$ implies inequalities $(1+y)\ln(1+y) \geq y$ and $|\ln(1-x)| \leq |\frac{-x}{1-x}| = \frac{x}{1-x}$ for $x \in (0,1)$. Thus for $x \in (0,1)$:

$$\frac{(1+y)\ln(1+y)}{|\ln(1-x)|} \geq \frac{y \cdot (1-x)}{x} = (1-x)\cdot \sum_{i = 0}^{n-2} x^i = 1-x^{n-1}$$

Finally for $x \in (0,1)$:

$$(1-x^{n-1})(1+x) = 1 - x^n + x (1 - x^{n-2}) > 1 - x^n - nx^{n-1}(1-x) $$

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