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Let $$M(N)=\sum_{n=1}^{N}\mu(n)$$

be a Mertens function.

According to the fact that $$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}$$

And fact that $\zeta(0)=-\frac{1}{2}$

Then i conclude that $$\lim_{N\to\infty}M(N)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{0}}=\frac{1}{\zeta(0)}=-2$$

Hence $M(N)$ is bounded ... But as we know it is not bounded.

Could you spot the mistake?

Regards.

mkultra
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    Unfortunately, the formula for $1/\zeta(s)$ only holds for $Re(s)>1$, so $s=0$ cannot be concluded. – Dietrich Burde May 22 '19 at 19:37
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    $\frac{1}{\zeta(s)}=\prod_p (1-p^{-s})=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}$ is true for $\Re(s) > 1$, the prime number theorem is that $\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}$ stays true for $\Re(s) \ge 1$, the Riemann hypothesis is that it holds for $\Re(s) > 1/2$, for $\Re(s) \le 1/2$, $\frac{1}{\zeta(s)}$ isn't given by a Dirichlet series anymore (see how it works for the analytic continuation of $\sum_{n=1}^\infty n^{-s}$), the functional equation is that it is a gamma factor times a Dirichlet series in $1-s$. – reuns May 22 '19 at 19:58
  • Yet another "interesting" function is $f(x)=\displaystyle\sum_{n=1}^{\infty}\mu(n)x^n$. Numerical experiments might make you think that $\displaystyle\lim_{x\to 1-0}f(x)=-2$, but actually the limit doesn't exist. – metamorphy May 22 '19 at 20:01
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    If $\lim_{x \to 0} f(e^{-x})$ exists or stays $O(x^{-\epsilon})$ then $\int_0^\infty x^{s-1} f(e^{-x})dx = \frac{\Gamma(s)}{\zeta(s)}$ is analytic for $\Re(s) > 0$ which we know isn't true since it has at pole at $\approx 1/2+14.13 i$. The same holds for $\frac{1}{\zeta(s)} = s \int_1^\infty M(x)x^{-s-1}dx$ if $M(x)$ is bounded or $O(x^{\epsilon})$. @metamorphy – reuns May 22 '19 at 22:00

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