# Sum of Prime Numbers

• June 9th 2010, 08:17 AM
chiph588@
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.
• June 10th 2010, 08:06 AM
Unbeatable0
I came up with two methods. Is one of them the one you used? (Happy)

Method 1:

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

Using $p_n \sim n \ln n$ :

$
\sum_{k=1}^n p_k \sim \sum_{k=1}^n k\ln k = \ln \prod_{k=1}^n k^k \sim
$

Now, using the result of this thread:

$
\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
$

Method 2:

Applying Stolz theorem

$
\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} =
$

Using $p_n \sim n \ln n$ :

$
= \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
$

Therefore $\sum_{k=1}^n p_k \sim \frac{1}{2}n^2\ln n$, as desired.
• June 10th 2010, 08:30 AM
chiph588@
Quote:

Originally Posted by Unbeatable0
I came up with two methods. Is one of them the one you used? (Happy)

Method 1:

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

Using $p_n \sim n \ln n$ :

$
\sum_{k=1}^n p_k \sim \sum_{k=1}^n n\ln n = \ln \prod_{k=1}^n k^k \sim
$

Now, using the result of this thread:

$
\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
$

Method 2:

Applying Stolz theorem

$
\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} =
$

Using $p_n \sim n \ln n$ :

$
= \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
$

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! (Clapping)

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.
• June 10th 2010, 08:49 AM
Unbeatable0
Quote:

Originally Posted by chiph588@
No I used a different method. Would you like to see it?

I like method 2! (Clapping)

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 (Wink)

$
\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
$
• June 10th 2010, 09:35 AM
chiph588@
Quote:

Originally Posted by Unbeatable0
I'm curious to see what method you used.

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

$
\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}
$

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