Does the following inequality always hold true?

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

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.

• Where does this come from? – Lee David Chung Lin Oct 6 '18 at 21:43
• I'm pretty sure it's from my head, after which I wrote it down here. – Joebloggs Oct 6 '18 at 22:25

$$\frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{p_n p_{p_n}} \leq \frac{\sum \limits_{j=n}^{n-1+p_n} p_j}{n \ln n p_{n \ln n}}$$ 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 $$1$$.

• 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. – Joebloggs Oct 8 '18 at 6:30