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The Big Omega function is defined as the number on non-distinct prime factors of an integer. I.e. $\Omega (2^a3^b...p^z)=a+b+...+z$, and the Liouville function is defined as $\lambda(x)=(-1)^{\Omega(x)}$.

It can also be shown that $$\frac{\zeta(2s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\lambda(k)}{k^s}$$

Now supposedly the statement $$ \sum\limits_{k=1}^{\infty}\frac{\lambda(k)}{k}=0 $$ Is equivalent to the prime number theorem. But surely it follows from $$\sum\limits_{k=1}^{\infty}\frac{\lambda(k)}{k}=\lim_{s\to1}\sum\limits_{k=1}^{\infty} \frac{\lambda(k)}{k^s}=\lim_{s\to1}{\frac{\zeta(2s)}{\zeta(s)}}=\frac{\zeta(2)}{\infty}=0$$ Thus we have proved the PNT? Something must be going wrong as surely no such easy proof of the PNT exists.

Elie Bergman
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  • "It can also be shown that $\frac{\zeta(2s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\lambda(k)}{k^s}$." Maybe the difficulty has "shifted" to the proof of this statement? – Daniel R Jan 15 '14 at 13:30
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    @DanielR For $\operatorname{Re} s > 1$, that's not hard to see using the Euler product. The difficulty would be showing $$\sum_{k=1}^\infty \frac{\lambda(k)}{k} = \lim_{s\to 1} \sum_{k=1}^\infty \frac{\lambda(k)}{k^s}$$ probably. Since the left sum doesn't converge absolutely, that may be much trickier than it looks. – Daniel Fischer Jan 15 '14 at 13:39
  • Yes you must be right. However I am still struggling to see an intuitive reason why we cannot force an equality. For each individual term, the difference between 1/k and 1/k^(1+epsilon) will become zero as epsilon becomes small. Since this is true for all terms, why can't we say that the two series converge to each other? – Elie Bergman Jan 15 '14 at 22:43
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    The individual terms converging doesn't suffice. Let $c_k = (3+(-1)^k)/2$, and consider the series $$S(\epsilon)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^{1+c_k\epsilon}}.$$ For $\epsilon\searrow0$, each term converges to $(-1)^{k-1}/k$, and the sum of the alternating harmonic series is $\log2$. But $$S(\epsilon)=(1-2^{-1-\epsilon})\zeta(1+\epsilon)-2^{-1-2\epsilon}\zeta(1+2 \epsilon),$$ and as $\epsilon\searrow0$, that is $\sim\frac{1}{4\epsilon}$, so tends to $+\infty$. – Daniel Fischer Jan 15 '14 at 23:36
  • Ok I see that the problem is a lot deeper in this case. But this is only to be expected from an equivalent PNT. – Elie Bergman Jan 15 '14 at 23:49

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