Show that $\sum_{n=d^k}^{d^{k+1} -1} x^{d(n+1)} \leq d^{(k+1)} x^{1 + d + \dots + d^{k+1}}$

Question

Let $x \in (0,1), d,k \in \{2,3,\dots\}$ I would like to show the following inequality: $$ \frac{1}{d^{(k+1)}}\sum_{n=d^k}^{d^{k+1} -1} x^{d(n+1)} \leq x^{1 + d + \dots + d^{k+1}}, $$ note how the left hand side is an average of $x^{d^{k+1}+d^k},\dots, d^{k+2}$ and that the power of $x$ on the right hands side lies somewhere in between the outer 2 most powers, more specifically it is one of the lower powers. Also the value of $x^n$ increases in $n$ as $n$ increases, this makes it very likely (imho) that this inequality indeed holds (I would expect some kind of convexity argument).

Answer 1

This cannot be true. The largest addend on the left is $x^{d^{k+1}+d}$, whose exponent is smaller than the exponent of $x$ on the right-hand side. Thus for fixed $d,k$, $x\searrow0$ leads to a contradiction.