Is there a real number $r$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$? Let $p_n$ denote the sequence of prime numbers, with $p_0=2$.
I'm looking for a real number $r$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$.
It's easy to show that $r>5$, with $\frac{2}{5^0}+\frac{3}{5^1}+\frac{5}{5^2}=2.8>e$.
I suppose it shouldn't be too hard to show that $r<6$.
So we know that $5<r<6$ (my observation shows that $r\approx5.7747052$).
But does it prove that there must be some value of $r\in\mathbb{R}$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$?

From number theory it is known that there are positive constants $C_1$ and $C_2$ such that one has 
$$C_1{n \ln n} < p_n < C_2{n \ln n}$$
Using this and the theory of power series you can show that there is such an $r$.
Look at the power series $f(x) = \sum_{n=1}^{\infty} p_n x^n$. Since $\lim_{n \rightarrow \infty} (n \ln n)^{-{1 \over n}} = 1$ (which you can prove by taking logs and using l'Hopital's rule), one also has that the radius of convergence of this power series is
$$\lim_{n \rightarrow \infty} (p_n)^{-{1 \over n}} = 1$$
So $f(x)$ is a continuous function on $(-1,1)$. Since the coefficients are all positive, one has $\lim_{x \rightarrow 1} f(x) = \infty$. You can see this for example by observing that
$\liminf_{x \rightarrow 1} f(x) $ is at least the limit of the sum of the first $n$ terms for any $n$, so
$\liminf_{x \rightarrow 1} f(x) \geq \sum_{k = 1}^n p_k $ Letting $n$ go to infinity shows the limit is in fact infinity.
So $f(x)$ is a continuous function on $[0,1)$ with $f(0) = 2$, and with $\lim_{x \rightarrow 1} f(x) = \infty$. So the intermediate value theorem says there is some $x$ for which $f(x) = e$. Letting $r = {1 \over x}$ gives what you are looking for.
