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Until a better answer appears. Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

It says that

$$s(p_n) \tilde \quad n^2 \log n /2.$$

where $p_n$ is the $n$-th prime.

Perhaps you want to look at the reference, and figure out if you can make the bound effective.

Until a better answer appears. Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

It says that

$$s(p_n) \sim \; n^2 \log n /2.$$

where $p_n$ is the $n$-th prime.

Perhaps you want to look at the reference, and figure out if you can make the bound effective.

Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

It says actually that

$$s(n) \tilde \quad n^2 \log n /2.$$

Until a better answer appears. Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

It says that

$$s(p_n) \tilde \quad n^2 \log n /2.$$

where $p_n$ is the $n$-th prime.

Perhaps you want to look at the reference, and figure out if you can make the bound effective.

Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

Here is a link:

http://mathworld.wolfram.com/PrimeSums.html

It says actually that

$$s(n) \tilde \quad n^2 \log n /2.$$