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

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- Jun 9th 2010, 08:17 AMchiph588@Sum of Prime Numbers
Show $\displaystyle \sum_{i=1}^n p_i \sim \frac12n^2\log n $ where $\displaystyle p_i $ is the $\displaystyle i^{th} $ prime number.

- Jun 10th 2010, 08:06 AMUnbeatable0
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 $\displaystyle p_n \sim n \ln n$ :

$\displaystyle

\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:

$\displaystyle

\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

$\displaystyle

\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 $\displaystyle p_n \sim n \ln n$ :

$\displaystyle

= \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 $\displaystyle \sum_{k=1}^n p_k \sim \frac{1}{2}n^2\ln n$, as desired. - Jun 10th 2010, 08:30 AMchiph588@
No I used a different method. Would you like to see it?

I like method 2! (Clapping)

On method 1, we know $\displaystyle p_n\sim n\log n $ but does that imply $\displaystyle \sum p_n\sim \sum n\log n $? I think perhaps that needs justification but I could be wrong. - Jun 10th 2010, 08:49 AMUnbeatable0
- Jun 10th 2010, 09:35 AMchiph588@