Let $k$ be a non-negative integer, $p$ be a prime number and consider the $\mathbb{Q}_p$-series $a(k,p) = \sum_{n=0}^{\infty} n^kp^n$ (it converges, since $n^kp^n \to 0$ in $\mathbb{Q}_p$ for obvious reasons). Prove that $a(k,p) \in \mathbb{Q}$ and that for any fixed $k$ the set $\{a(k,p): p - \mbox{prime}\}$ is infinite.
I am not aware of a general criterion to verify whether a series in $\mathbb{Q}_p$ is in $\mathbb{Q}$. Wolfram Alpha gave a really implicit closed form of this series; on the other hand I see that using derivatives suitably could work, but did not reach anything proper.
Any help appreciated!
Sketch: You can compute $a(k,p)$ directly using the geometric series and its derivations:
Let $$f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ so that $a(0,p)=f(p)=\frac{1}{1-p}$ is rational.
Now $f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{x}\sum_{n=0}^\infty nx^{n}$, which means $a(1,p)=pf'(p)$ is also rational and so on.
In general, for the function $a(k,x)=\sum_{n=0}^\infty n^k x^n$, we have $$\frac{d}{dx} a(k,x)=\sum_{n=0}^\infty n^{k+1} x^{n-1}=\frac{1}{x}\sum_{n=0}^\infty n^{k+1} x^{n}=\frac{a(k+1,x)}{x}$$
Edit: Well $a(0,p)=\{\frac{1}{1-p}:p\text{ is prime}\}$ is an infinite set. Now $a(1,p)=p\frac{d}{dp}(0,p)=\frac{p}{(1-p)^2}$ thus $\{a(1,p):p\text{ is prime}\}$ is also infinite.
Now suppose the set $\{a(k,p):p\text{ is prime}\}$ is infinite and prove by induction that $\{a(k+1,p):p\text{ is prime}\}$ is also infinite.