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Let $p_n$ be the $n^\text{th}$ prime and $g_n$ be the $n^\text{th}$ prime gap ($p_{n+1} - p_n$). Calculating the serie up to $n = 10^6$ it seems that

$$\sum_{n=1}^\infty \frac{n}{{p_n}^{g_n}} < 1$$

Could this be true and is there a way to prove it?

(This is a follow up question to this, although perhaps a harder nut to crack.)

Marijn
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  • How did you calculate the series up to $10^6$? Because on my machine the series up to $10^5$ took about 20 minutes to calculate – Dominic Michaelis Sep 11 '15 at 19:39
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    Sounds reasonable since the gap is soon $2$ or more and the $n$-th prime is greater than a constant times $n\log n$. – André Nicolas Sep 11 '15 at 19:43
  • @DominicMichaelis I have a file with the first million primes and use JavaScript to go through it and do the calculations. Took about 8 seconds. – Marijn Sep 11 '15 at 19:56
  • @Marijn So I guess you do it numerically? machine precision will make a lot of problems in such calculations ... – Dominic Michaelis Sep 11 '15 at 19:59
  • @DominicMichaelis You're right, it's by no means a perfect way. – Marijn Sep 11 '15 at 20:06
  • @DominicMichaelis Machine precision is a much smaller issue than people think in such simple calculations. Floating point accuracy is $10^{-16}$ and if we assume random errors (sometimes adding and sometimes subtracting) it accumulates roughly as $\sqrt{N}10^{-16}$. The worst case is $N10^{-16}$. Both are really small for $N=10^6$ as here. – Winther Sep 11 '15 at 20:13
  • @Winther maybe I used the wrong word for it, I mean how many significant digits we have, because for example if the first time is surprisingly big (for example like 10^100) an the other terms are all 1, numerically the series should be constant 10^100 shouldnt it? – Dominic Michaelis Sep 11 '15 at 20:19
  • @DominicMichaelis I see. Yes, that is a valid point in general since $1.0+10^{-20} = 1.0$ If there were like $10^{20}$ numbers with size $10^{-20}$ that would indeed be a problem (one can partly get around this by summing smartly, i.e. adding smaller numbers in groups to get bigger number and add them). – Winther Sep 11 '15 at 21:22

3 Answers3

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Note that $g_n\ge 2$ except at the beginning. Also, by Rosser's Theorem we have $p_n\gt n\log n$. Then the result follows by Comparison with the convergent series $\sum_2^\infty \frac{1}{n\log^2 n}$. (One needs much less than Rosser's Theorem, the Prime Number Theorem is enough.)

André Nicolas
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We have $g_n\geq 2$ so $\frac{1}{p_n^{g_n}} \leq \frac{1}{p_n^2}$. Further we have $p_n> n\log n$ for $n\geq 1$ by the prime number theorem (or Rosser's theorem to be precice). This implies that

$$S \leq \sum_{n=1}^{k}\frac{n}{p_n^{g_n}} + \sum_{n=k}^\infty\frac{1}{n(\log n)^2} \leq \sum_{n=1}^{k}\frac{n}{p_n^{g_n}} + \frac{1}{\log(k)}$$

since $ \sum_{n=k}^\infty\frac{1}{n(\log n)^2} \leq \int_k^\infty \frac{dx}{x\log^2(x)} = \frac{1}{\log(k)}$. Computing the first sum above numerically we get

$$S \leq 1.355$$

for $k=10$. To push the bound below $1$ seems to require more precise estimates.

Winther
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    Calculating with mathematica the first $10^5$ terms give 0.971055 as result, but with the $\frac{1}{\log(k)}$ bound even this isn't enough as $\log(10^5)=5\cdot \log(10)\approx 0.0868589$ – Dominic Michaelis Sep 11 '15 at 20:04
  • @DominicMichaelis Yes one needs at least $10^{15}$ terms computed by the computer to be able to use these estimates to prove it. It seems unlikely that one is able to push the bound below $1$ without some really precise estimates. – Winther Sep 11 '15 at 20:09
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By Chebyshev's theorem (the weak version of the PNT) it follows that $p_n \geq C n \log n$ for some constant $C$ close to $1$, for any $n$ big enough. Since $g_n\geq 2$ for any $n\geq 2$ and the series $$\sum_{n\geq 2}\frac{1}{n\log^2(n)}$$ is convergent by Cauchy's condensation test, your series is convergent, too.

Is the exact value of the original series really relevant for some purpose?

Jack D'Aurizio
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