Showing $\sum_{x=1}^\infty x^j \rho^x \leq \frac{j!}{(1- \rho)^{j+1}}$, where $\rho \in (0,1)$ and $j\in \mathbb{N}_0$ I have trouble to understand why the following inequality holds:
$$\sum_{x=1}^\infty x^j \rho^x \leq \frac{j!}{(1- \rho)^{j+1}}$$
where $\rho \in (0,1)$ and $j\in \mathbb{N}_0$.
I have tried to link it to the geometric series and and taking the derivative, but it gets too complicated with higher derivatives and I got lost.
 A: $$\frac{j!}{(1-\rho)^{j+1}}=\frac{\partial^j}{\partial\rho^j}\frac1{1-\rho}=\frac{\partial^j}{\partial\rho^j}\sum_{n=0}^\infty\rho^n=\sum_{n=0}^\infty (n+1)\cdots(n+j)\rho^n\geqslant\sum_{n=0}^\infty n^j\rho^n.$$
A: My best illustration for the j = 0 case for this problem follows, which elaborates on @Idontgetit's comment about the geometric series.  I have added an update for cases where $j > 0$ below.
The inequality can be rewritten so that 1 is the only term on the right-hand side.  Since both the numerator and denominator are guaranteed to be greater than zero, the inequality direction can be preserved:
$$\sum_{x= 1}^{\infty} x^j \rho^x \le \frac{j!}{(1- \rho)^{j+1}}$$
$$\frac{(1- \rho)^{j+1}}{j!} \sum_{x= 1}^{\infty} x^j \rho^x \le 1$$
The j = 0 case gives the geometric series multiplied by $\rho$:
$$\sum_{x= 1}^{\infty} x^j \rho^x = \rho \sum_{x= 1}^{\infty} 1* \rho^{x-1} = \frac{\rho}{1 - \rho}$$
Substituting into the inequality, we get:
$$\frac{(1- \rho)^{0+1}}{0!} \frac{\rho}{1 - \rho} \le 1$$
We can cancel the $(1 - \rho)$ term from numerator and denominator:
$$ \frac{\rho}{j!} \le 1$$
Examining the above, it clearly reduces to $$ \rho \le 1$$ for the case of j = 0.
Update, 7/10/22: cases where $j > 0$.
This informal illustration uses the identity noted by aschepler given here
as well as the idea that $A_j(\rho) \le j!$ for $\rho \in (0,1)$.
First, note that
$$\sum_{x= 1}^{\infty} x^j \rho^x = \sum_{x= 0}^{\infty} x^j \rho^x$$
because the first term on the right-hand side must be zero.  Next, by the Eulerian identity expression noted by @aschepler, we have:
$$\sum_{x= 0}^{\infty} x^j \rho^x = \frac{\rho * A_j(\rho)}{(1- \rho)^{j+1}}$$
where
$$A_j(\rho) = \sum_{m = 0}^{j-1} A(j,m)\rho^{m}$$
and
$$A(j,m) =  \sum_{k = 0}^{m+1}(-1)^k\binom{j+1}{k}(m + 1 - k)^j$$
Because $A_j(1) = j!$ for $j > 0$ and $A_j(\rho)$ is strictly increasing on $\rho \in (0,1)$, $A_j(\rho) \le j!$.  The function is strictly increasing because each term in $A_j(\rho)$ is multiplied by $\rho$ to some power.  That $A_j(1) = j!$ is a known property. Plugging $j!$ into the expression yields:
$$\sum_{x= 1}^{\infty} x^j \rho^x \le  \frac{\rho * j!}{(1- \rho)^{j+1}} \le \frac{j!}{(1- \rho)^{j+1}}$$
All terms on both sides of the inequality cancel, being greater than 0 in every case, yielding:
$$ \rho \le 1$$
for $j > 0$.
A: Notice that $$ S_j(\rho) := \sum_{n \ge 1} n^j \rho^n = \sum_{n \ge 1} \rho^n \int_{z = 0}^n j z^{j-1} \mathrm{d}z = \sum_{n\ge 1} \int_{z = 0}^n j \rho^n z^{j-1}\mathrm{d}z. $$ The expression being summed and integrated above is non-negative, and so we can exchange the order of the summation and integration by Tonelli's theorem. This gives $$ \frac{S_j(\rho)}{j} =  \int_{z = 0}^\infty z^{j-1} \sum_{n \ge \lceil z \rceil } \rho^n \mathrm{d}z = \int_{z = 0}^\infty \frac{z^{j-1} \rho^{\lceil z\rceil} }{1-\rho} \mathrm{d}z.$$
Since $\rho <1, \rho^{\lceil z \rceil} \le \rho^z,$ giving $$ \frac{S_j(\rho) (1-\rho)}{j} \le \int_{z= 0}^{\infty} z^{j-1} \rho^z \mathrm{d}z = \frac{(j-1)! }{(\log (1/\rho))^j},$$ where we have used the definition and properties of the Gamma function. Notice, btw, that this is tight up to a factor of $\rho$ since $\rho^{\lceil z \rceil} \ge \rho^{z+1}$.
Finally, since $\log x \le x-1,$ it follows for $x \in (0,1)$ that $\frac{1}{-\log x} \le \frac{1}{1-x}.$ This tells us that $$ (1-\rho) \frac{S_j(\rho)}{j} \le \frac{(j-1)!}{(1-\rho)^{j}} \iff S_j(\rho) \le \frac{j!}{(1-\rho)^{j+1}}. \\.$$
