It is well known that

$$ \sum_{n\geq1} \frac{1}{n(n+1)}=1, $$


$$ \sum_{n=1}^{N} \frac{1}{n(n+1)}=1-\frac{1}{N-1}. $$

But what of

$$ \sum_{n\geq1} \frac{1}{p_n(p_n+1)}, $$

where $p_n$ denotes the n$^{\rm th}$ prime? More generally, does the sum

$$ S(k)=\sum_{n\geq1} \frac{1}{p_n^k(p_n+1)^k} $$

converge? If so, to what?


Since most comments consist of "it converges using the comparison test", I would like to specify that I am looking for the value of these sums, not a statement about whether they converge or not.

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    $\begingroup$ Convergence is quite trivial, because $\forall n\, p_n>n$. $\endgroup$ – TZakrevskiy Jun 22 '18 at 8:44
  • $\begingroup$ math.stackexchange.com/questions/2826155/… , greetings. $\endgroup$ – Peter Szilas Jun 22 '18 at 8:49
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    $\begingroup$ If you have a convergent sum of positive terms, then every subsum (which takes only a subset of all summands) converges to a smaller number. $\endgroup$ – Crostul Jun 22 '18 at 8:55
  • $\begingroup$ @TZakrevskiy If it is so trivial, could you please say what it converges to? Wolfram Alpha returns "Computation time exceeded", instead of a solution (wolframalpha.com/input/…). If it would be trivial I assume WA would give an answer quite rapidly. $\endgroup$ – Klangen Jun 22 '18 at 9:07
  • $\begingroup$ @PreservedFruit What I mean is that it is trivial to show that the series converges to a finite value $S(1)\in \Bbb R$; finding exact value of $S(1)$ is, unfortunately, not that trivial. $\endgroup$ – TZakrevskiy Jun 22 '18 at 9:11

Let $P(s)=\sum_p\frac{1}{p^s}$ be the prime zeta function.

$$\sum_{p}\frac{1}{p(p+1)} = \sum_{m\geq 2}(-1)^m P(m) = \sum_{m\geq 2}(-1)^m\sum_{k\geq 1}\frac{\mu(k)}{k}\log\zeta(km) $$ allows an efficient numerical evaluation of the LHS, which turns out to be pretty close to $\frac{1}{3}$ ($\approx 0.33023$). I do not believe there is any "nice" closed form for such series or for the similar Meissel-Mertens constant, but the same approach applies to the series $\sum_{p}\frac{1}{p^s(p+1)^s}$.

A similar technique is to notice that over the interval $\left[0,\frac{1}{5}\right]$ the approximation $$\frac{x^2}{x+1}\approx \log(1+x^2)-\log(1+x^3)+\frac{3}{2}\log(1+x^4) $$ is extremely accurate, hence $$ \sum_{p}\frac{1}{p(p+1)}\approx \frac{1}{4}+\sum_{p\geq 5}\left[\log\left(1+\frac{1}{p^2}\right)-\log\left(1+\frac{1}{p^3}\right)+\frac{3}{2}\log\left(1+\frac{1}{p^4}\right)\right] $$ where the RHS can be computed from Euler's product, as $$\frac{1}{4}-2\log\pi+\log(2^4 3^6 7)+\frac{1}{2}\log(2\cdot 3\cdot 7)-\frac{1}{2}\log(5\cdot 17\cdot 41)-\log(17\cdot 41)-\log\zeta(3)\approx\\ \approx \color{green}{0.330}616\ldots $$ Proving that $\sum_{p\text{ prime}}\frac{1}{p(p+1)}<\frac{1}{3}$ is actually a very nice exercise, thank you for the suggestion.

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  • $\begingroup$ Hi Jack. Thanks a lot for your excellent response. The first step (i.e., that $\sum_{p}\frac{1}{p(p+1)} = \sum_{m\geq 2}(-1)^m P(m)$ is not immediately obvious to me. Could you please explain it? $\endgroup$ – Klangen Jun 25 '18 at 8:50
  • $\begingroup$ @PreservedFruit: $\frac{1}{p(p+1)}$ is a geometric series, $\frac{1}{p^2}-\frac{1}{p^3}+\frac{1}{p^4}-\frac{1}{p^5}+\ldots$ $\endgroup$ – Jack D'Aurizio Jun 25 '18 at 12:31
  • $\begingroup$ Indeed :) Thank you! $\endgroup$ – Klangen Jun 25 '18 at 12:51

A value with a precision of roughly 683 decimal digits is given in http://oeis.org/A179119

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Correct me if wrong:

A proof of the 'trivial':

$a_k= \dfrac{1}{k(k+1)}$, $k=1,2,3,...$

$b_k = \dfrac{1}{p_n(p_n+1)}$ for

$k=p_n, n=1,2,3...$

$b_k =0$ for $k\not = p_n$, $n=1,2, 3,..$

Then $b_k \le a_k$ , $k \in \mathbb{N}.$


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