Computing $\sum_{n=1}^{\infty} \frac{p_n}{n!}$ where $p_n$ is the $n^{th}$ prime number.

Question

I have come across this sum today and have not been able to reach someone to discuss it. This converges by the ratio test for all real values of $x$.

How would one go about computing the sum ?

$\sum_{n=1}^{\infty} \frac{p_n}{n!}$ where $p_n$ is the $n^{th}$ prime number.

Answer 1

We can compute it quite precisely by bounding the tail.

We have \begin{align*} \sum_{n=1}^\infty \frac{p_n}{n!} = \sum_{n=1}^k \frac{p_n}{n!} + \sum_{n=k+1}^{\infty} \frac{p_n}{n!} \end{align*} and since $p_n < n^2$ for all $n>1$, we have $$ \sum_{n=k+1}^{\infty} \frac{p_n}{n!} < \sum_{n=k+1}^{\infty} \frac{n^2}{n!} < \sum_{n=k+1}^{\infty} \frac{1}{2^n} = \frac{1}{2^k} $$ using the fact that $n^2 2^n < n!$ for $n \ge 8$.

So, with $k\ge 8$, we have $$ \sum_{n=1}^k \frac{p_n}{n!} < \sum_{n=1}^\infty \frac{p_n}{n!} < \sum_{n=1}^k \frac{p_n}{n!} +\frac{1}{2^k} $$

Thus summing up the first 100 terms gives us a value of the sum accurate to (at least) $30$ decimal digits: the sum is approximately $$ 4.73863870268621999271779470348 $$ with all digits correct.

(Much tighter bounds could be used: the first 100 terms gives us much more accuracy, but this is easy, and $30$ digits is enough for a lot of purposes.