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Math Help - Sum of Prime Numbers

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Sum of Prime Numbers

    Show  \sum_{i=1}^n p_i \sim \frac12n^2\log n where  p_i is the  i^{th} prime number.
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  2. #2
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    I came up with two methods. Is one of them the one you used?

    Method 1:

    (I prefer this one, as it "finds" the asymptote)

    Using p_n \sim n \ln n :

    <br />
\sum_{k=1}^n p_k \sim \sum_{k=1}^n k\ln k = \ln \prod_{k=1}^n k^k \sim<br />

    Now, using the result of this thread:

    <br />
\sim \ln C + \frac{1}{2}\left(n^2+n+\frac{1}{6}\right)\ln n - \frac{1}{4}n^2 \sim \frac{1}{2}n^2\ln n<br />



    Method 2:

    Applying Stolz theorem

    <br />
\lim_{n\rightarrow \infty}\frac{n^2 \ln n}{\sum_{k=1}^n p_k} = \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{p_n} =<br />

    Using p_n \sim n \ln n :

    <br />
 = \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{n \ln n} = \lim_{n\rightarrow \infty}\frac{n \ln(1+\frac{1}{n-1}) + 2\ln(n-1) - \frac{\ln(n-1)}{n}}{\ln n} = 2<br />

    Therefore \sum_{k=1}^n p_k \sim \frac{1}{2}n^2\ln n, as desired.
    Last edited by Unbeatable0; June 10th 2010 at 10:15 AM. Reason: Typo in summation
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    I came up with two methods. Is one of them the one you used?

    Method 1:

    (I prefer this one, as it "finds" the asymptote)

    Using p_n \sim n \ln n :

    <br />
\sum_{k=1}^n p_k \sim \sum_{k=1}^n n\ln n = \ln \prod_{k=1}^n k^k \sim<br />

    Now, using the result of this thread:

    <br />
\sim \ln C + \frac{1}{2}\left(n^2+n+\frac{1}{6}\right)\ln n - \frac{1}{4}n^2 \sim \frac{1}{2}n^2\ln n<br />



    Method 2:

    Applying Stolz theorem

    <br />
\lim_{n\rightarrow \infty}\frac{n^2 \ln n}{\sum_{k=1}^n p_k} = \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{p_n} =<br />

    Using p_n \sim n \ln n :

    <br />
 = \lim_{n\rightarrow \infty}\frac{n^2 \ln n - (n-1)^2 \ln (n-1)}{n \ln n} = \lim_{n\rightarrow \infty}\frac{n \ln(1+\frac{1}{n-1}) + 2\ln(n-1) - \frac{\ln(n-1)}{n}}{\ln n} = 2<br />

    Therefore \sum_{k=1}^n p_k \sim \frac{1}{2}n^2\ln n, as desired.
    No I used a different method. Would you like to see it?

    I like method 2!

    On method 1, we know  p_n\sim n\log n but does that imply  \sum p_n\sim \sum n\log n ? I think perhaps that needs justification but I could be wrong.
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  4. #4
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    Quote Originally Posted by chiph588@ View Post
    No I used a different method. Would you like to see it?

    I like method 2!

    On method 1, we know  p_n\sim n\log n but does that imply  \sum p_n\sim \sum n\log n ? I think perhaps that needs justification but I could be wrong.

    I'm curious to see what method you used.

    Regarding the justification, it's again an application of Stolz theorem

    <br />
\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^n p_k}{\sum_{k=1}^n k \ln k} = \lim_{n\rightarrow  \infty}\frac{p_n}{n \ln  n} = 1<br />
    Last edited by Unbeatable0; June 10th 2010 at 10:16 AM. Reason: Typo in summation
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    I'm curious to see what method you used.

    Regarding the justification, it's again an application of Stolz theorem

    <br />
\displaystyle{\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^n p_k}{\sum_{k=1}^n n \ln n} = \lim_{n\rightarrow  \infty}\frac{p_n}{n \ln  n} = 1}<br />
    My solution was more complicated since I was unsure of the asymptotic justification you provided me.

    Spoiler:


    Using partial summation  \displaystyle{\sum_{p\leq x} p = x\cdot\pi(x)-\int_2^x\pi(t)dt}  \displaystyle{= x\left(\frac{x}{\log x}+O\left(xe^{-a\sqrt{\log x}}\right)\right)-\int_2^x \left(\frac{t}{\log t}+O\left(te^{-a\sqrt{\log t}}\right)\right)dt}

     \displaystyle{= \frac{x^2}{\log x}-\int_2^x \frac{t}{\log t}dt+O\left(x^2e^{-a\sqrt{\log x}}\right)+O\left(\int_2^xte^{-a\sqrt{\log t}}dt\right)}

    Next I divided what I have above by  \frac12\frac{x^2}{\log x} and showed the limit was one. I omit my work because it's pretty straightforward.

    Thus  \displaystyle{\sum_{p\leq x}p\sim\frac12\frac{x^2}{\log x}\implies\sum_{i=1}^np_i=\sum_{p\leq p_n}p\sim\frac12\frac{p_n^2}{\log{p_n}} \sim \frac12\frac{n^2\log^2n}{\log n+\log\log n}  \sim\frac12n^2\log n} .

    Last edited by chiph588@; July 17th 2010 at 05:03 PM.
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