# Is it possible to assign a value to the sum of primes?

It is possible, by means of zeta function regularization and the Ramanujan summation method, to assign a finite value to the sum of the natural numbers (here $n \to \infty$) : $$1 + 2 + 3 + 4 + \cdots + n \mathrel{\unicode{x201c;}{=}\unicode{x201d;}} - \frac{1}{12} .$$ Is it also possible to assign a value to the sum of primes, $$2 + 3 + 5 + 7 + 11 + \cdots + p_{n}$$ ($n \to \infty$) by using any summation method for divergent series?

This question is inspired by a question on quora.

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You can if you approximate $p_n\rightarrow n\ln(n)$ –  user1708 Nov 22 '11 at 18:08
If you can, it won't be as nice or have as much meaning as it does for the zeta function. Your first sum can be more concretely written as $\zeta(-1)=\frac{1}{12}$. However, the prime zeta function cannot be analytically continued to the left of the imaginary axis. –  Eric Naslund Nov 22 '11 at 18:18
@anon: Yes, I am aware of that. I wrote that regularization and summability methods assigns finite values to infinite, divergent series. –  Max Muller Nov 22 '11 at 18:26
@howdy :how? (text) –  Max Muller Nov 22 '11 at 18:33
@Max: Note that $p_n\sim n\log n$ by the prime number theorem. You can zeta-regularize the divergent sum $\sum_{n=1}^\infty n\log n$ by evaluating $-\zeta'(-1)=\log A-1/12$, where $A$ is the Glaisher-Kinkelin constant. So it's an answer to something similar to your question. –  anon Nov 22 '11 at 18:48

Fröberg shows in his paper that the prime zeta function

$$P(s)=\sum_{p\in \mathbb P} \frac1{p^s}=\sum_{k=1}^\infty \frac{\mu(k)}{k}\log\zeta(ks)$$

where $\mu(k)$ and $\zeta(s)$ are respectively the Möbius and Riemann functions, cannot be analytically continued to the left half-plane, $\Re\,s\leq 0$ (in particular, we cannot give a reasonable evaluation of $P(-1)$), due to the clustering of poles along the imaginary axis arising from the nontrivial zeros of the Riemann $\zeta$ function.

Note the nasty-looking left edges in both plots above.

This result is originally due to Landau and Walfisz. See the linked papers for more details.

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Notes: The formula is basically the infinite version of Mobius inversion. And the $P(-1)$ can't be defined because a dense line of poles forms a blockade against any hope of analytic continuation. –  anon Nov 22 '11 at 18:20
Ok, but perhaps there is a different method by means of which it can be done, right? Analytic continuation is only one of many ways to sum divergent series. –  Max Muller Nov 22 '11 at 18:22
@J.M.: Here is a short proof: In your formula above, notice that when $s=\frac{1}{k}$ for some squarefree $k$, then we have a $\log\zeta(1)$ term appear in the sum, which means it is a singular point. This sequence $\frac{1}{k}$ for $k$ squarefree is then a sequence of singularities which converges to zero. –  Eric Naslund Nov 22 '11 at 18:28
Also note that this result is originally due to Landau and Walfisz. –  Eric Naslund Nov 22 '11 at 18:31
Thanks, Eric and anon, for the added details. –  Guess who it is. Nov 22 '11 at 18:39