0
$\begingroup$

How can you evaluate $\sum_{n = 1}^\infty \left( \frac{1}{\sum (n)} \right)$ where $\sum (n)$ is the sum of the first $n$ prime numbers? I was able to find numerical bounds by running code to give a lower bound of $1.02345$, and I found an upper bound by taking the reciprocal of the Prime Sum Function approximation $\frac{\ln(n)\cdot n^2}{2}$ and evaluating a sum from there, but I would like to figure out how to evaluate this exactly.

$\endgroup$
3
  • $\begingroup$ Most randomly chosen infinite series cannot be evaluated in closed form. I don't see any reason why there would be a closed form for this series. $\endgroup$ Nov 23 '20 at 8:47
  • $\begingroup$ Thank you, that is very interesting. However, as this sum isn't random and sums the primes, shouldn't there be a closed form? I'm curious:) $\endgroup$ Nov 23 '20 at 9:05
  • $\begingroup$ Most sums don't have closed form. $\endgroup$ Nov 23 '20 at 9:43
0
$\begingroup$

Exactly would be very difficult.

You want to compute $$S=\sum_{n = 1}^\infty \frac{1}{a_n} \qquad \text{where} \qquad a_n=\sum _{i=1}^n p_i$$ What I should do it to write it as $$S=\sum_{n = 1}^p \frac{1}{a_n}+\sum_{n = p+1}^\infty \frac{1}{a_n}$$

Concerning $$s_p=\sum_{n = 1}^p \frac{1}{a_n}$$ they form the sequence $$\left\{\frac{1}{2},\frac{7}{10},\frac{4}{5},\frac{73}{85},\frac{2129}{2380},\frac {89669}{97580},\frac{2649191}{2829820},\frac{29545361}{31128020},\frac{7464160 3}{77820050}\right\}$$ For the second summation, I should use as an approximation of $a_n$ $$\frac{n^2}{2}\left[\ln n + \ln\ln n - \frac{3}{2} + \frac{\ln\ln n}{\ln n} - \frac{5}{2\ln n} - \frac{\ln^2\ln n}{2\ln^2 n} + \frac{7\ln\ln n}{2\ln^2 n} - \frac{29}{4\ln^2 n} + \cdots\right]$$ as proposed by Sinha in $2011$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.