0

I was toying around with some prime number related series (trying to generalize some results from a puzzle) and came across this one:

$$\sum_{p \text{ prime}} \frac{1}{p^2+p}$$

Is there any reasonable way to calculate this? If not, can you get some nice bounds on it? (above and below would be nice. I know it lies between $0.3$ and $0.4$, and running it through a program actually tells me it is about $0.33$)

Sofía Garbayo
  • 38
  • 7
  • 1
    We can only calculate this sum numerically. With which precision would you be content ? – Peter May 23 '20 at 06:29
  • 3
    FYI, for some bounds on the sum of $\frac{1}{p(p-1)}$, which is similar to your $\frac{1}{p(p+1)}$, please check Prime Telescoping Series. – John Omielan May 23 '20 at 06:36
  • 3
    Check this link –  May 23 '20 at 06:57
  • @ metamorphy,the sum is between $1/6=0.166666666667$ and $0.644934066848$ ,and the sum itself is $0.26149721284$ which is true by the bounds I've given. –  May 23 '20 at 07:09
  • @user715522: Did you check this? BTW, here's my PARI/GP session: $$\begin{array}{l} \texttt{foo(n)=sumdiv(n,d,if(d==1,0,moebius(n/d)d(-1)^d))} \\texttt{logzeta(a,n)={my(r=zeta(n));forprime(p=2,a,r=1-p^(-n));return(log(r))}} \\texttt{goo(a)={my(r=suminf(n=2,foo(n)/nlogzeta(a,n)));forprime(p=2,a,r+=1/p/(p+1));return(r)}} \\texttt{\p100} \\texttt{goo(50)} \end{array}$$ – metamorphy May 23 '20 at 07:15
  • Oh, you are right. I calculated those bounds for half the sum actually, which was my original problem. I'll edit that – Sofía Garbayo May 23 '20 at 07:17

1 Answers1

0

I can't give you an answer, but.

If we consider $$ \sum_{p \leq x} \frac{1}{p(p + 1)} = \sum_{p \leq x} \left( \frac{1}{p} - \frac{1}{p + 1} \right), $$ the question amounts to determining the constant $A$ appearing in the asymptotic formula of the form $\sim \log \log x + A + o(1)$ for the sum $$ \sum_{p \leq x}\frac{1}{p + 1}. $$

I don't think A takes an easy form.

sozinhozinho
  • 66