Let Pi denote the i-th prime number. Is there any formula for expressing S=∑i=1mPi. We know that there are around Pmln⁡(Pm) prime numbers less than or equal to Pm. So, we have: S≤m×Pm≤Pm2ln⁡(Pm) I want to know, if there is a better bound for S, in the litrature.

Summation by parts gives ∑p≤np=∑k=1n(π(k)−π(k−1))k =nπ(n)+∑k=1n−1π(k)(k−(k+1)) =nπ(n)−∑k=1n−1π(k)...(1) We have that π(k)=klog⁡(k)(1+O(1log⁡(k))) and so using the Euler-Maclaurin Sum Formula, we get that ∑k=1n−1π(k)=12n2log⁡(n)+O(n2log⁡(n)2)...(2) Therefore, we get ∑p≤np=12n2log⁡(n)+O(n2log⁡(n)2)...(3) Setting n=Pm should give you a closer estimate.

Expert Answer to "Let Pi denote the i-th prime number. Is..."

Cariglinom5

Answered question

2021-12-03

Let

P_{i}

denote the i-th prime number. Is there any formula for expressing

S = \sum_{i = 1}^{m} P_{i}

.
We know that there are around

\frac{P_{m}}{\ln (P_{m})}

prime numbers less than or equal to

P_{m}

. So, we have:

S \leq m \times P_{m} \leq \frac{P_{m}^{2}}{\ln (P_{m})}

I want to know, if there is a better bound for S, in the litrature.

Answer & Explanation

Kathleen Ashton

Beginner2021-12-04Added 15 answers

Summation by parts gives

\sum_{p \leq n} p = \sum_{k = 1}^{n} (π (k) - π (k - 1)) k

= n π (n) + \sum_{k = 1}^{n - 1} π (k) (k - (k + 1))

= n π (n) - \sum_{k = 1}^{n - 1} π (k)

...(1)
We have that

π (k) = \frac{k}{\log (k)} (1 + O (\frac{1}{\log (k)}))

and so using the Euler-Maclaurin Sum Formula, we get that

\sum_{k = 1}^{n - 1} π (k) = \frac{1}{2} \frac{n^{2}}{\log (n)} + O (\frac{n^{2}}{{\log (n)}^{2}})

...(2)
Therefore, we get

\sum_{p \leq n} p = \frac{1}{2} \frac{n^{2}}{\log (n)} + O (\frac{n^{2}}{{\log (n)}^{2}})

...(3)
Setting

n = P_{m}

should give you a closer estimate.

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