<s>[INST] How to show the monotonicity of this function? Thanks for your help. I want to show $f(x)=\displaystyle \frac{1-x^{n+1}(n+2)+(n+1)x^{n+2}}{(1-x^{n+1})(1-x)}$ (strictly) increases w.r.t. $x$ for $x>0$. Here, $n=1,2,3,\dots.$. So far, I can only show $f(x)$ is increasing when $x>1$. Can anyone give me a hint to complete the proof?   Thanks a lot. 
 [/INST] I really like this question.
Applying division and then partial fractions, you can write your function as
$$n + 1 + \frac{1}{1-x} - \frac{n+1}{1-x^{n+1}}.$$
Since $f$ is continuous except at $x=1$, and its limit exists there (I'll leave it to you to show this), to show it is strictly increasing we need to prove that its derivative
$$\frac{1}{(1-x)^2} - \frac{(n+1)^2x^n}{(1-x^{n+1})^2}$$
is positive almost everywhere. 
First, by AM-GM, for $x\in (0,1)\cup (1,\infty)$
$$\frac{1+x+x^2+\ldots+x^n}{1+n} > x^{n/2}.$$
Therefore
\begin{align*}
\frac{1-x^{n+1}}{1-x} &> (1+n)x^{n/2}\\
\frac{(1-x^{n+1})^2}{(1-x)^2} &> (1+n)^2x^n\\
\frac{1}{(1-x)^2} &> \frac{(1+n)^2x^n}{(1-x^{n+1})^2},
\end{align*}
so
$$\frac{1}{(1-x)^2} - \frac{(1+n)^2x^n}{(1-x^{n+1})^2} > 0.$$
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