proving monotonicity of sequence involving logarithms. I am repeatedly getting sequences that involve natural logarithms and I have to apply Dini's theorem to those functions (see below): $$\dfrac{1}{n}\,\dfrac{1}{\alpha x^n}\,\ln\left(1-\alpha x^n\right)$$ where $\,0< \alpha ,\, x <1\,$ and $\,n=1,2,\ldots$
Is there a way to prove that a sequence like this is monotonic? 
 A: Let $f(n)=\frac{1}{αnx^n}\ln{[1-αx^n]}$. Then 
\begin{align}\frac{\partial}{\partial n}f(n)&=-\frac{1+n\ln{[x]}}{αn^2x^n}\ln{[1-αx^n]}-\frac{αx^n\ln{[x]}}{αnx^n(1-αx^n)}\\[0.3cm]&=-\frac{(1+n\ln[x])(1-αx^n)\ln{[1-αx^n]}+αnx^n\ln[x]}{αn^2x^n(1-αx^n)}\end{align} Now all the terms in the denominator are positive - $(1-αx^n)>0$ by assumption - and similarly both summands in the numerator are negative - $1+n\ln[x]<0$ for sufficiently large $n$ and $\ln[x]<0$. So, with the $-$ in front the derivative becomes non-negative  for all $n$.
A: Different way:
\begin{aligned}
&\dfrac{1}{n+1}\,\dfrac{1}{\alpha x^{n+1}}\,\ln\left(1-\alpha x^{n+1}\right)-\dfrac{1}{n}\,\dfrac{1}{\alpha x^n}\,\ln\left(1-\alpha x^n\right)\\
&>\dfrac{1}{n+1}\,\dfrac{1}{\alpha x^{n+1}}\,\ln\left(1-\alpha x^{n+1}\right)-\dfrac{1}{\color{red}{n+1}}\,\dfrac{1}{\alpha x^n}\,\ln\left(1-\alpha x^n\right)\\
&=\dfrac{1}{\alpha(n+1)x^{n+1}}\left(\ln\left(1-\alpha x^{n+1}\right)-x\ln\left(1-\alpha x^n\right)\right)\\
&>\dfrac{1}{\alpha(n+1)x^{n+1}}\left(\ln\left(1-\alpha x^{n+1}\right)-\color{red}1\ln\left(1-\alpha x^n\right)\right)\\
&>\dfrac{1}{\alpha(n+1)x^{n+1}}\left(\ln\left(1-\alpha x^{\color{red}n}\right)-\ln\left(1-\alpha x^n\right)\right)\\
&=0\\
\end{aligned}
