Some estimation a series by integral Let $a \in (0,1)$. Does there exist a constant $C>0$ or function $C(a)>0$, which may be a function of $a$ but not $k$, such that
$$
\sum_{n=1}^\infty a^n n^k \leq C(a) \int_0^\infty a^x x^k dx
$$
for all $k \in \mathbb N$?
 A: Recalling the Mellin transform 
$$ F(s)=\int_{0}^{\infty} x^{s-1} f(x) dx. $$
Our integral is nothing but the Mellin transform of the function $a^x=e^{\ln(a)x}$. Now, if $a \in (0,1) $ then $\ln(a) < 0 $ and we have
$$ \int_{0}^{\infty} a^x x^{k} dx = \int_{0}^{\infty} x^{k} e^{\ln(a)x} dx= \int_{0}^{\infty} x^{k} e^{-\ln(\frac{1}{a})x} dx = \frac{\Gamma(k+1)}{(-\ln(a))^{k+1}}. $$
The last integral can be evaluate by using the change of variables $ t=\ln\left (\frac{1}{a} \right)x $ and the gamma function
$$ \Gamma( s+1 ) = \int_{0}^{\infty} t^{s}e^{-t}dt. $$
Note: Here are some numerical values for the sum $S$ and the integral $I$ which show they are almost equal 
$$ (k,a)=(3,1/2) \implies S = 26.00000000,\quad I = 25.99258100 $$
$$ (k,a)=(10,1/7) \implies S = 2395.590764,\quad I = 2395.589595 $$
A: Let:
\begin{align}
a&\in (0,1)\\\
k&\in\mathbb{N}\\
f_k&\equiv \sum_{n=1}^\infty a^n n^k\\\
g_k&\equiv \int_{x=0}^\infty a^x x^k dx.
\end{align}
Then:
\begin{align}
f_0 &= \frac{a}{1-a}\\\ \\
f_k &= a\frac{df_{k-1}}{da}\\\ \\
g_k &= \frac{(-1)^{k+1}k!}{\ln(a)^{k+1}}.
\end{align}
For $g_k$ we're lucky enough to have an explicit solution (as pointed out by Mhenni). The recurrence relation for $f_k$ together with the initial condition uniquely define the entire series $f_k\ \forall\ k\in\mathbb{N}$:
\begin{align}
f_1 &= \left(\frac{a}{1-a}\right) + \left(\frac{a}{1-a}\right)^2 = \frac{a}{(a-1)^2}\\
f_2 &= \left(\frac{a}{1-a}\right) + 3\left(\frac{a}{1-a}\right)^2 + 2\left(\frac{a}{1-a}\right)^3 = \frac{a(a+1)}{(1-a)^3}\\
f_3 &= \left(\frac{a}{1-a}\right) + 7\left(\frac{a}{1-a}\right)^2 + 12\left(\frac{a}{1-a}\right)^3 + 6\left(\frac{a}{1-a}\right)^4=\frac{a(a^2+1+4a)}{(a-1)^4}\\
&\ldots
\end{align}
Each of the terms in the $f_k$ are positive, so $f_{k+1}>f_k\ \forall\ k\in \mathbb{N}$.
Using the above, one can define the ratio $C_k \equiv f_k/g_k$.  All of the various $C_k$ are finite functions of $a$, which vary smoothly in the interval $a\in(0,1)$.  However, if you want to define the inequality $f_k\leq Cg_k\ \forall k\in\mathbb{N}\ \forall a\in(0,1)$ for some $C\in \mathbb{R}$ you would need $C=\max(C_k)$ where the maximization is over both $k$ and $a$.  I'm not sure if such a $C$ exists or not.
A: Given $n,k\in\mathbb{N}$ and  $a\in(0,1)$, when $x\in[n,n+1]$, $a^xx^k>a^{n+1}n^k$. It follows that
$$\int_0^\infty a^xx^kdx>\sum_{n=1}^\infty\int_n^{n+1}a^xx^kdx>\sum_{n=1}^\infty a^{n+1}n^k=a\cdot\sum_{n=1}^\infty a^nn^k.$$
Edit: The argument above shows that $C(a)$ can be chosen as $a^{-1}$. Actually $C$ cannot be independent of $a$. From $\sum_{n=1}^\infty a^nn^k>a$ and $\int_0^\infty a^x x^kdx=\frac{k!}{(\log a^{-1})^{k+1}}$ we know that 
$$R(a,k):=\frac{\sum_{n=1}^\infty a^nn^k}{\int_0^\infty a^x x^kdx}>\frac{a\cdot(\log a^{-1})^{k+1}}{k!}.$$
Let $a=e^{-k}$, then
$$R(e^{-k},k)>\frac{k^{k+1}}{e^k\cdot k!}.$$ 
Due to Stirling' formula, $\lim_{k\to\infty}R(e^{-k},k)=\infty$. 
