Upper bound for infinite sum Let $x \in (0,1)$, I need to find an upper bound (as good as possible) for the series
$$\sum_{n=1}^{\infty}x^n  n^k,$$
where $k$ is a natural number larger or equal than $2$, i.e., $k=2,3,4,\dots$.
My try: I know how to do it for $k=1$:
$$x^n(n+1)=(x^{n+1})'$$
and I suppose that something similar works for the rest of $k\ge2$. For instance, if $n>k$
$$\frac{d^k}{dx^k}(x^n)=n\cdot(n-1)\cdot(n-2)\cdots (n-k)x^{n-k-1}\le n^kx^{n-k-1}.$$
But, then I am getting confused and I cannot get a clean way to finish this.
Edit: Also, I was told that for $k=4$, it holds that $$\sum_{n=1}^\infty x^n n^4\le \frac{24}{x^3(1-x)}$$ but I don't know how to verify this.
Edit2: So, assuming that the above is correct, is there a way to prove the conjecture that
$$\sum_{n=1}^\infty x^n n^k\le \frac{k!}{x^{k-1}(1-x)}?$$
 A: 1. The proposed inequality cannot be true for $k > 0$, because
$$ \sum_{n=0}^{\infty} n^k x^n \sim \frac{k!}{(1-x)^{k+1}} \quad \text{as} \quad x \to 1^- $$
and hence the sum will blow up at a speed of $(1-x)^{-k-1}$.
2. What is true is
$$ \frac{k!x^k}{(1-x)^{k+1}} \leq  \sum_{n=0}^{\infty} n^k x^n \leq \frac{k!}{(1-x)^{k+1}} $$
This will be particularly useful when $x$ is close to $1$, and also shows that the upper bound correctly captures the leading term in the Laurent expansion of the sum about $x = 1$.
Or, extending @Sungjin Kim's answer, we have the following identities
\begin{align*}
\sum_{n=0}^{\infty} n^k x^n
&= \sum_{j=0}^{k} \biggl\{ {k \atop j} \biggr\}  \frac{j! x^j}{(1-x)^{j+1}} \\
&= \sum_{j=0}^{k} (-1)^{j-k} \biggl\{ {k + 1 \atop j + 1} \biggr\} \frac{j!}{(1-x)^{j+1}},
\end{align*}
where $\bigl\{{k \atop j}\bigr\}$ is the Stirling numbers of the second kind. Again, note that the second line is precisely the Laurent expansion of the series about $x = 1$, and the leading term is $\frac{k!}{(1-x)^{k+1}}$.
Yet another representation is
$$ \sum_{n=0}^{\infty} n^k x^n = \frac{x A_n(x)}{(1-x)^{n+1}}, $$
where $A_n(x)$ is the Eulerian polynomials. In short, there are ways to expand the infinite series in the form of a combinatorial sum, but I am not sure if this helps in your situation.
A: We can write the sum as a finite sum using Stirling numbers of the second kind. We have
$$
\sum_{n=1}^{\infty}x^nn^k=\frac1{1-x}\sum_{m=1}^k S(k,m)m! \left(\frac x{1-x}\right)^m.
$$
We obtain an upper bound by applying an upper bound of Stirling numbers ($S(k,m)\leq \frac12 \binom km m^{k-m}$ if $k\ge 2$, $m\le k-1$. If $k\geq 2$,
$$
\frac1{1-x}\sum_{m=1}^k S(k,m)m! \left(\frac x{1-x}\right)^m\leq \frac1{1-x}\sum_{m=1}^{k-1} \frac12 \binom km m! m^{k-m}\left(\frac x{1-x}\right)^m+\frac1{1-x} k! \left(\frac x{1-x}\right)^k.
$$
