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June 10th, 2014, 02:32 PM   #11
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A better explanation, though a bit rough : Consider the zeta function $$\zeta(s) = \sum_{n \geq 1} \frac1{n^s}$$ This doesn't look right. What about $\zeta(-1)$? Clearly it diverges. Indeed, the definition is only valid if $\mathcal{R} s > 1$. So, naturally, one would ask how does the zeta function "behave" around $\mathcal{R} s =1$. Uh, too complicated, let us work with $s = 1$.

Well, it diverges. Indeed, $\zeta(1)$ is just a harmonic series and by integral test it diverges. But there is more to it. We have denominator running through all of $\Bbb N$, correct? So let us regroup terms and use fundamental theorem of arithmetic to get some relation with the primes in $\Bbb N$. For the sake of it, we use this untidy ambiguous notation just to make things cleaner, and keep in mind that we are actually thinking of finitely many terms and then setting the limit tend arbitrarily large

$$\left ( 1 + \frac12 + \frac13 + \frac14 + \frac15 + \cdots \right )\cdot \left( 1 - \frac12 \right) = 1 + \frac13 + \frac15 + \frac17 + \frac19 + \cdots$$

This was fun. Why not do it again with $3$?

$$\left ( 1 + \frac12 + \frac13 + \frac14 + \frac15 + \cdots \right )\cdot \left( 1 - \frac12 \right) \left (1 - \frac13 \right) = 1 + \frac15 + \frac17 + \frac1{10} + \cdots$$

So continue this until we have only $1$ left at the RHS.

$$\left ( 1 + \frac12 + \frac13 + \cdots \right) \prod_p \left ( 1 - \frac1p \right) = 1$$
$$\Rightarrow \left ( 1 + \frac12 + \frac13 + \cdots \right) = \prod_p \left ( 1 - \frac1p \right )^{-1}$$

Now note that this immediately gives the infinitude of primes. As otherwise, there would be finitely many terms in the product, i.e., that would converge but that'd contradict the divergence of the harmonic series.

So, you see, a simple analysis around $s = 1$ gives a a very interesting data about primes. So again one can naturally ask what happens if $s$ is complex, i.e, $\mathcal{R} s = 1$?. Alas, we have just crossed our boundaries of simple math and looking at this essentially is quite dangerous without a good deal of hard analysis backed up. It's a matter of fact that you can actually proof that $\zeta$ has no zeros at the line $\Re s = 1$, say, if you approach it from the above (since we don't have our formula working at that particular line). And, indeed, it can actually give you a lot more than infinitude. In particular, one can prove the prime number theorem $$\pi(x) \sim x/\log(x)$$ in a similar form, which is actually much stronger than infinitude.

So, definitely, one would like to ask more. One would hope (though in vain... or is it so?) that $\zeta$ could have a more domain of convergence, perhaps one could understand many open problems regarding primes.

Well, it's not quite true that $\zeta$ is not defined in the domain $\mathcal{R} s < 1$. True, the series doesn't converge there but there is a particular nature of this zeta function that can be mimicked, namely, analyticity. Differentiability. One can actually introduce a function throughout $\Bbb C/\{0\}$ which is analytic and match with the values of zeta for $\mathcal{R} s > 1$. This is accomplished by delicate results of complex analysis and is much more complicated that you can think it is. Indeed, the most intriguing region for the zeta is $\mathcal{R} s = 1/2$ and it is conjectured that all of zeta zeros are situated there, and the implications to the theory of prime numbers would be massive. Putting it into a highly non literal way : you can know where exactly the primes are inside $\Bbb Z$ once you know the situation of the zeros of $\zeta$, which is what Riemann hypothesis is about.

Enough babbling!
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June 10th, 2014, 07:18 PM   #12
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Isn't the Riemann Zeta function analytic on $C$\{1}?
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June 11th, 2014, 12:25 AM   #13
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Sure. But the $p$-series isn't, which is just Riemann zeta for $\mathcal{R} s > 1$.
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