Induction proof using mean value theorem

Question

Let $\alpha\in(0,1)$. Use the mean-value theorem to show that ln$(1-\alpha^k)>-\alpha^k/(1-\alpha)$ for all $k\in \mathbb{N}$.

So I know that the mean value theorem is $\frac{f(b)-f(a)}{b-a}=f'(c)$ for some $c\in (a,b)$.

So I am thinking that the intervals to use the MVT should be $[k,k+1]$.

So for the base case, $k=1$, $[1,2]$ implies $\frac{\ln(1-\alpha^2)-\ln(1-\alpha)}{(1-\alpha^2)-(1-\alpha)}=\frac{1}{1-\alpha^c}$ where $c\in(01)$

So $\frac{\ln(1-\alpha^2)-\ln(1-\alpha)}{(2)-(1)}=\ln(1-\alpha^2)-\ln(1-\alpha)=\ln(1-\alpha^2)-\ln(1-\alpha)=\frac{1}{1-\alpha^c}$

I know that $\frac{1}{1-\alpha^c}>\frac{1}{1-\alpha}>0>\frac{-\alpha}{1-\alpha}$

Now is where I am stuck. So I need to show that $\ln(1-\alpha^2)-\ln(1-\alpha)=\ln(1-\alpha)$

or $\ln(1-\alpha^2)-\ln(1-\alpha)<\ln(1-\alpha)$, but how? Or am I completely wrong?

Answer 1

$\forall k \in \mathbb{N}_{+}$ by applying the MVT with $f:x->ln(1-x)$ on $I=[\alpha^k;1]$ $\exists c \in [\alpha^k;1]$ such as:

$\frac{ln(1)-ln(1-\alpha^k)}{1-\alpha^k}=\frac{-1}{1-c}$.

so:

$ln(1-\alpha^k) = \frac{1}{1-c} - \frac{\alpha^k}{1-c} $

$\forall x \in I $: $\frac{1}{1-\alpha}\geq \frac{1}{1-\alpha^k}\geq\frac{1}{1-x}$ and $ \frac{1}{1-x}\geq0$. This is true in particular for $c$ so:

$ln(1-\alpha^k) \geq - \frac{\alpha^k}{1-\alpha}$

The case $k=0$ is simple.

Answer 2

Here is a proof using Taylor series. Suppose $x \in (0,1)$, then $\log(1-x) = - \sum_{n=1}^\infty {x^n \over k} > - \sum_{n=1}^\infty x^n$.

Hence $\log(1-\alpha^k) > - \sum_{n=1}^\infty (\alpha^k)^n = - {\alpha^k \over 1- \alpha^k} \ge - {\alpha^k \over 1- \alpha} $.