Is there a real number $r$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$?

Question

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 $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$?

Answer 1

Such an $r$ does exist. The series is a Laurent series for a function continuous on a portion of $(0,\infty)$ that approaches 2 as $r$ approaches $\infty$ and approaches infinity as $r$ approaches some real number greater than or equal to 0. The intermediate value theorem shows the function takes on any positive value greater than 2.

To see that the series converges, we may quote the bound on $p_n$ (see this article) $$p_n<n\log(n)+n\log(\log(n))$$ thus $p_n<(n+1)(n+2)$ for large enough $n$. We have that $$\frac{1}{(1-\frac{1}{r})^3}=\sum_{n}{\frac{(n+1)(n+2)}{r^n}}$$ which bounds our series and converges for all $r>1$, so $\sum_n{\frac{p_n}{r^n}}$ converges for all $r>1$.

If a power series converges in some neighborhood of a point, then it converges to a continuous (even holomorphic) function. What matters is how quickly the sequence increases. $a_n=n!$ is one example which wouldn't converge for any value of $r$.

Answer 2

Write $f(x) = \sum_k p_k x^k$. Assume first that the series for $f(1/4)$ converges. Then the function $f(x)$ is defined and continuous on $[0,1/4)$. You've shown $f(1/5) > e$, and clearly $f(0) = 2$. By the intermediate value theorem, there must be some $c \in (0,1/5)$ such that $f(c) = e$. You can take $r = 1/c$.

It remains to show that $f(1/4)$ is finite. The convergence of this series will follow from the root test if we can establish, for instance, that $p_n < 3^n$. I imagine that there is an easier way to show this, but at minimum this follows by induction from Bertrand's postulate.

I don't believe the general arguments you gave are sufficient unless the series is shown to converge at some point $d$ for which $f(d) \geq e$, which will inevitably involve some estimate of the growth of $p_n$.

Answer 3

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.