Mixing
Asymptotic result for sum of every second prime number
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In this question, I asked about the convergence of $hatalpha$, the intercept of the best fit line joining all prime numbers. I stumbled on this result today where it was shown that $sum_{k=1}^n p_ksimfrac12n^2ln n$ and similar results for prime powers.
Are there asymptotic results for $sum_{k=1}^n p_{2k}$ and $sum_{k=1}^n p_{2k-1}$?
If we denote $pi_{2k}(x)=sum_{p_{2k}}1$ then $sum_{k=1}^n p_{2k}=int_1^x t,dpi_{2k}(t)$. Now $pi_{2k}(x)=frac12pi(x)pm{0,1}$ depending on circumstance and since $pi(x)simint_2^xfrac{du}{ln u}$, considering the simplest case of $pm0$, $$sum_{k=1}^n p_{2k}sim frac12int_1^xfrac{t}{ln t},dt.$$ How should I continue; is there a quicker way to obtain the asymptotics?
number-theory convergence prime-numbers asymptotics
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
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|
show 2 more comments
$begingroup$
In this question, I asked about the convergence of $hatalpha$, the intercept of the best fit line joining all prime numbers. I stumbled on this result today where it was shown that $sum_{k=1}^n p_ksimfrac12n^2ln n$ and similar results for prime powers.
Are there asymptotic results for $sum_{k=1}^n p_{2k}$ and $sum_{k=1}^n p_{2k-1}$?
If we denote $pi_{2k}(x)=sum_{p_{2k}}1$ then $sum_{k=1}^n p_{2k}=int_1^x t,dpi_{2k}(t)$. Now $pi_{2k}(x)=frac12pi(x)pm{0,1}$ depending on circumstance and since $pi(x)simint_2^xfrac{du}{ln u}$, considering the simplest case of $pm0$, $$sum_{k=1}^n p_{2k}sim frac12int_1^xfrac{t}{ln t},dt.$$ How should I continue; is there a quicker way to obtain the asymptotics?
number-theory convergence prime-numbers asymptotics
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
$endgroup$
$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
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– reuns
Jan 26 at 12:37
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@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
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– TheSimpliFire
Jan 26 at 12:39
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41
|
show 2 more comments
$begingroup$
In this question, I asked about the convergence of $hatalpha$, the intercept of the best fit line joining all prime numbers. I stumbled on this result today where it was shown that $sum_{k=1}^n p_ksimfrac12n^2ln n$ and similar results for prime powers.
Are there asymptotic results for $sum_{k=1}^n p_{2k}$ and $sum_{k=1}^n p_{2k-1}$?
If we denote $pi_{2k}(x)=sum_{p_{2k}}1$ then $sum_{k=1}^n p_{2k}=int_1^x t,dpi_{2k}(t)$. Now $pi_{2k}(x)=frac12pi(x)pm{0,1}$ depending on circumstance and since $pi(x)simint_2^xfrac{du}{ln u}$, considering the simplest case of $pm0$, $$sum_{k=1}^n p_{2k}sim frac12int_1^xfrac{t}{ln t},dt.$$ How should I continue; is there a quicker way to obtain the asymptotics?
number-theory convergence prime-numbers asymptotics
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
$endgroup$
In this question, I asked about the convergence of $hatalpha$, the intercept of the best fit line joining all prime numbers. I stumbled on this result today where it was shown that $sum_{k=1}^n p_ksimfrac12n^2ln n$ and similar results for prime powers.
Are there asymptotic results for $sum_{k=1}^n p_{2k}$ and $sum_{k=1}^n p_{2k-1}$?
If we denote $pi_{2k}(x)=sum_{p_{2k}}1$ then $sum_{k=1}^n p_{2k}=int_1^x t,dpi_{2k}(t)$. Now $pi_{2k}(x)=frac12pi(x)pm{0,1}$ depending on circumstance and since $pi(x)simint_2^xfrac{du}{ln u}$, considering the simplest case of $pm0$, $$sum_{k=1}^n p_{2k}sim frac12int_1^xfrac{t}{ln t},dt.$$ How should I continue; is there a quicker way to obtain the asymptotics?
number-theory convergence prime-numbers asymptotics
number-theory convergence prime-numbers asymptotics
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
edited Jan 26 at 11:37
TheSimpliFire
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
asked Jan 26 at 11:26
TheSimpliFireTheSimpliFire
12.8k62461
12.8k62461
$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
$endgroup$
– reuns
Jan 26 at 12:37
$begingroup$
@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
$endgroup$
– TheSimpliFire
Jan 26 at 12:39
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41
|
show 2 more comments
$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
$endgroup$
– reuns
Jan 26 at 12:37
$begingroup$
@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
$endgroup$
– TheSimpliFire
Jan 26 at 12:39
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41
$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
$endgroup$
– reuns
Jan 26 at 12:37
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
$endgroup$
– reuns
Jan 26 at 12:37
$begingroup$
@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
$endgroup$
– TheSimpliFire
Jan 26 at 12:39
$begingroup$
@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
$endgroup$
– TheSimpliFire
Jan 26 at 12:39
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41
|
show 2 more comments
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$begingroup$
You can use integration by parts to find the last integral is asymptotically $frac{1}{2}frac{x^2}{ln x}$. Then use the fact $xsim 2nlog n$.
$endgroup$
– Wojowu
Jan 26 at 11:38
$begingroup$
@Wojowu Right, would you mind posting an answer for the asymptotics of the sums of $p_{2k}$ and $p_{2k-1}$? It'd be useful for future reference!
$endgroup$
– TheSimpliFire
Jan 26 at 12:30
$begingroup$
I don't see what you mean. You can't find the asymptotic of $sum_{k le n} p_{2k}$ without knowing that $p_k = k log k+o(klog k)$ which is the PNT. From there it is obvious that $sum_{k le n} p_{2k} sim sum_{k le n} 2k log(2k) sim n^2 log n$
$endgroup$
– reuns
Jan 26 at 12:37
$begingroup$
@reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$
$endgroup$
– TheSimpliFire
Jan 26 at 12:39
$begingroup$
$sum_{k le n} k log k ge log(n^{1-epsilon}) sum_{n^{1-epsilon} le k le n} k = log(n^{1-epsilon}) ( n(n+1)/2 -n^{1-epsilon}(n^{1-epsilon}+1)/2)$
$endgroup$
– reuns
Jan 26 at 12:41