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$$ 0\lt \frac{\sum_{i=n}^{n + P_n - 1} P_i}{P_n \cdot P _{P_n}} \leq 1 $$

Or is there a lower bound bigger than zero? Which I believe not to be the case.

Some basic examples are as follows:

$(1)$ Numerator:
$n = 4, P_n = P_4 = 7, P_{n + P_n - 1} = P_{4 + P_4 - 1} = ... P_{4 + 7 - 1} = P_{10} = 29$, i.e. Sigma $P_4$ to $P_{10}$ = Sigma all primes between $7$ & $29$, which equals $119$.

Denominator:
$P_n \cdot P_{P_n} = P_4 \cdot P_{P_4} = 7 \cdot P_7 = 7\cdot17 = 119$ also, in this case.

Fraction $= 119/119 = 1$

$(2)$ Numerator: $n = 133, P_n = P_{133} = 751, P_{n + P_n - 1} = ... P_{133 + P_133 - 1} = P_{133 + 751 - 1} = P_{883} = 6863$, i.e. Sigma $P_{133}$ to $P_{883}$ = Sigma all primes between $751$ & $6863$, which equals $2772829$.

Denominator:
$P_n \cdot P_{P_n} = P_{133} \cdot P_{P_{133}} = 751 \cdot P_{751} = ... 751\cdot 5701 = 4281451$.

Fraction $= 2772829/4281451 = 0.64763...$

$(3)$ Numerator:
$n = 684, P_n = P_{684} = 5113, P_{n + P_n - 1} = ... P_{684 + P_684 - 1} = P_{684 + 5113 - 1} = P_{5796} = 57149$, i.e. Sigma $P_{684}$ to $P_{5796}$ = Sigma all primes between $5113$ & $57149$, which equals $154944253$.

Denominator:
$P_n \cdot P_{P_n} = P_{684} \cdot P_{P_684} = 5113 * P_{5113} = ... 5113\cdot49789 = 254571157$.

Fraction $= 154944253/254571157 = 0.60864...$

That's it.

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  • $\begingroup$ Where does this come from? $\endgroup$ – Lee David Chung Lin Oct 6 '18 at 21:43
  • $\begingroup$ I'm pretty sure it's from my head, after which I wrote it down here. $\endgroup$ – Joebloggs Oct 6 '18 at 22:25
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$\frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{p_n p_{p_n}} \leq \frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{n \ln n p_{n \ln n}}$ since $ p_n \geq n \ln n$

$\frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{n \ln n p_{n \ln n}} \leq \frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{n \ln n* n \ln n *\ln (n \ln n)} = \frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{n^2 \ln^2 n (\ln n +\ln \ln n)}$

$\sum \limits_{j=n}^{n-1+p_n} p_j \leq \int \limits_{n}^{n+p_n} p_j dj \leq \int \limits_{n}^{n+p_n} \frac{6}{5} j \ln j dj$ becuase $p_n \leq n(\ln n+\ln \ln n)$ and $\lim \limits_{n \to \infty} \frac{\ln \ln n}{\ln n}=0$

So $\int \limits_{n}^{n+p_n} \frac{6}{5} j \ln j dj \leq \int \limits_{1}^{n+p_n} \frac{6}{5} j \ln j dj \leq 0.3+ 1.2(0.5n^2 (\ln n+\ln \ln n+1)^2 \ln(n+1.2 n \ln n))$

The limit $ \lim \limits_{n\to \infty} \frac{0.3+ 1.2(0.5n^2 (\ln n+\ln \ln n+1)^2 \ln(n+1.2 n \ln n))}{n^2 \ln^2 n (\ln n +\ln \ln n)} = 0.6$.

Checking small cases conclude the proof.

So your sum approach $0.6$ when $n$ approach infinity, actually your sum limits is $0.5$, but this need a lot of work to prove, but at least we know that the sum always less or equal to $1$.

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  • 1
    $\begingroup$ Wow. I wouldn't have been able to prove that as you did as the whole thing was just a hunch, but you've done a great job here. Thanks so much, it must have taken a lot of time to do this and I really appreciate it Ahmad. $\endgroup$ – Joebloggs Oct 8 '18 at 6:30

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