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. 
 A: 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
$$\frac{1}{(1-x)^2} - \frac{(1+n)^2x^n}{(1-x^{n+1})^2} > 0.$$
A: Here's a solution that doesn't use calculus. As user7530 points out, we may simplify the expression to $(n+1) + \frac{1}{1-x} - \frac{n+1}{1-x^{n+1} } $.
Now, for $ a> b$, we want to show that (the constant term $(n+1)$ cancels out)
$$  \frac{1}{1-a} - \frac{n+1}{1-a^{n+1} }> \frac{1}{1-b} - \frac{n+1}{1-b^{n+1} }$$
This is equivalent to 
$$\frac{ (a-b) } { (1-a)(1-b) } > \frac{ (n+1) (a^{n+1}-b^{n+1} )} { (1-a^{n+1})(1-b^{n+1})} $$
which is equivalent to
$$ \frac{ (1-a^{n+1}) } {(1-a)} \times \frac{ (1-b^{n+1}) } { (1-b) } > (n+1) \frac{ a^{n+1} - b^{n+1} } { a-b}$$
which is equivalent to
$$( 1 + a + a^2 + \ldots + a^n) ( b^n + b^{n-1} + \ldots + 1) > (n+1) (a^n + a^{n-1}b + \ldots + ab^{n-1} + b^n) $$
which is true by the rearrangement inequality.

In hindsight, the last statement is similar to 7530's AM-GM.
