Sum of Prime Numbers

**chiph588@** · June 9th 2010, 08:17 AM

Show $\sum_{i=1}^n p_i \sim \frac12n^2\log n$ where $p_i$ is the $i^{th}$ prime number.

**Unbeatable0** · June 10th 2010, 08:06 AM

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 $p_n \sim n \ln n$ :

$ \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.

**chiph588@** · June 10th 2010, 08:30 AM

Originally Posted by Unbeatable0

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 $p_n \sim n \ln n$ :

$ \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!

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.

**Unbeatable0** · June 10th 2010, 08:49 AM

Originally Posted by chiph588@

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

$ \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 $

**chiph588@** · June 10th 2010, 09:35 AM

Originally Posted by Unbeatable0

I'm curious to see what method you used.

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

$ \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.

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