Why is $\sum_{k=1}^\infty k(-x)^{k-1}(1+(-x)^{k})^n$ decreasing for $x \in [0, 1]$? Let $n$ be a positive integer, and define $G_n(x)$ by
$$G_n(x) = \sum_{k=1}^\infty k(-x)^{k-1}(1+(-x)^{k})^n.$$
Notice,
$$G_n(x) = (1-x)^n - 2x(1+x^2)^n+3x^2(1-x^3)^n - \cdots$$

Why is $G_n(x)$ decreasing on the interval $[0,1]$?

This is a result that is needed to fill a hole in the argument of a very helpful answer to another question I posed. It took me quite a while for me to come up with a (WolframAlpha assisted) proof of it.

Is there a simpler justification for $G_n(x)$ to be decreasing than the answer I am posting below?

 A: Let $r_k(x) = kx^{k-1}(1+x^{k})^n$ and $h_k(x) = kx^{k-1}(1-x^{k})^n$. We have
$$G_n(x) = (1-x)^n + \sum_{k=1}^\infty (-r_{2k}(x) + h_{2k+1}(x)).$$
Notice that $(1-x)^n$ is decreasing on $[0, 1]$. So it is sufficient to show that $-r_{2k}(x) + h_{2k+1}(x)$ is decreasing, which we do by showing its derivative is negative on $[0,1]$. After a bit of simplifying the derivatives, we get
$$r_k'(x) = kx^{k-2}(1+x^k)^{n-1}(k - 1 + (kn + k - 1)x^{k-1})$$
$$h_k'(x) = kx^{k-2}(1-x^k)^{n-1}(k - 1 - (kn + k - 1)x^{k-1})$$
We will in fact show that $r_{k}'(x) > h_{k+1}'(x)$ on $[0,1]$ for $k =2$ and for $k \geq 4$. Dividing $r_k'(x)$ and $h_{k+1}'(x)$ by $x^{k-2}$, our goal is to prove this:
$$\begin{aligned}
&k(1+x^k)^{n-1}(k - 1 + (kn + k - 1)x^{k-1}) \\
&> (k+1)x(1-x^{k+1})^{n-1}(k - (kn + n + k)x^{k+1})
\end{aligned} \quad (*)$$
Note that for $x \in (0, 1)$, the left hand side of (*) is increasing in $n$, and the right hand side is decreasing in $n$, and so it is sufficient to prove ($*$) for $n=1$:
$$k(k-1 + (2k-1)x^{k-1}) > (k+1)x(k-(2k+1)x^{k+1}) $$
We will in fact show that for $k \geq 4$ the following stronger inequality holds
$$k(k-1 + (k-1)x^{k-1}) > (k+1)x(k-kx^{k+1})$$
or
$$k(k-1)(1+x^{k-1}) > (k+1)kx(1-x^{k+1}).$$
Removing the $(1+x^{k-1})$ term, we need only show the  following stronger inequaility
$$\frac{k-1}{k+1} > x(1-x^{k+1}).$$
Let $p(x) = x(1-x^{k+1})$. Then $p'(x) = 1 - (k+2)x^{k+1}$, and so $p(x)$ is maximized for $x = (k+2)^{-1/(k+1)}$. So we need only show $\frac{k-1}{k+1} >p((k+2)^{-1/(k+1)})$, i.e. show
$$\frac{k-1}{k+1} > \frac{1 - 1/(k+2)}{(k+2)^{1/(k+1)}}.$$
WolframAlpha says that this inequality is true for $k > 3.5921$, and so it is true for $k \geq 4$.
Going back to before we removed the $(1+x^{k-1})$ term and plugging in $k=2$, we now only need to show
$$(1+x) > 3x(1-x^3),$$
which is easy to check by finding the minimum of $f(x) = 1-2x+3x^4$.
