# Does $\sum_{n=1}^{\infty} \frac{p_n}{n!}$ converge?

Introduction

I was so fascinated the first time I saw Eulers Number $e$ as the sum of the infinite series over $\frac{1}{n!}$ that I began playing on variants of it, learning a lot on the way.

However when I started playing around with primes things became tricky, I found: $$\sum_{n=1}^{\infty} \frac{p_n}{n!}=\frac{2}{1!}+\frac{3}{2!}+\frac{5}{3!}+\dots = 4.73863870268..$$

It intrigued me when I noticed that it appeared to converge as I know about the divergence of the sums of reciprocal primes. I couldn't find this in any literature and couldn't find a closed form.

I have ran simulations and computed this for fun to hundreds of decimal places however I have ploted a graph for $\frac{p_{n+1}}{(n+1)!} - \frac{p_n}{n!}$, $n$ ranging from $1$ to $9$

Plot

I have checked with a program if every next difference is smaller than the previous and this holds up to $n=1000$ (with the idea that if this holds on forever it will eventually become $0$ thus the series will converge to a constant value)

Question #1: Does it converge?

Question #2: Is this something known (probably), and if yes, where could I read up on it?

• To prove convergence it is enough to exploit the ridiculous bound $1<p_n<e^n$. – Jack D'Aurizio Jan 28 '17 at 20:17
• By the way, although you mention that $$\sum_{n=1}^\infty \frac{1}{p_n} \to \infty,$$ the following infinite series is actually convergent: $$\sum_{n=1}^\infty\frac{(-1)^n}{p_n}\approx -0.2696063519.$$ This was proven in $1971$, and I thought that you might be interested. It is also unknown whether or not $$\sum_{n=1}^\infty \frac{n(-1)^n}{p_n}$$ is convergent or divergent. – Mr Pie Feb 16 '18 at 11:29
• @user477343 what was proven in 1971 ? Convergence of $\sum_{n=1}^\infty\frac{(-1)^n}{p_n}$ is trivial by the alternating series theorem. – Gabriel Romon Apr 22 '18 at 11:23

Yes, it converges. As a hint, the prime number theorem gives us an asymptotic estimate $$p_n \sim n \log n.$$
• @Ilhan Instead of the prime number theorem (complicated) you can show that $\sum_n \frac{1}{p_n}$ diverges (using that $\log \zeta(s)= -\sum_n \log(1-p_n^{-s})$), which means that $\lim_{n \to \infty} \frac{p_n}{n^2} = 0$, so that $\sum_n \frac{p_n}{n!}$ converges – reuns Jan 28 '17 at 20:26
• Given $p_n < (1+\varepsilon)n\log{n}$, for $\forall n > N$, you can play around $$\sum_{n}\frac{p_n}{n!}<\sum_{n=1}^{N}\frac{p_n}{n!}+(1+\varepsilon)\sum_{n>N}\frac{\log{n}}{(n-1)!}$$ and reduce this further given that $$\lim_{n\rightarrow \infty} \frac{\log{n}}{n-1}=0$$ – rtybase Jan 28 '17 at 20:39