One such example which I can think of is that Mertens conjecture was proven false. This states that Mertens function which is given by $$ M(x)=\sum_{n\leq x}\mu(n) $$ satisfies $\vert M(x)\vert\leq \sqrt{x}$. I should note that $\mu(n)$ is the Mobius function in number theory which is $0$ unless $n$ is square-free in that case it is given by $$ \mu(p_1p_2...p_k)=(-1)^k $$ One reason that the disproof of this conjecture is disappointing is that Mertens conjecture implies the Riemann hypothesis, RH is equivalent to the statement that $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$. Thus, we expect the square root cancelation in the summation; however, the disappointing part of Mertens’ conjecture being false is that as much as we expect RH to be true, if one were to attempt to prove RH by showing $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$, then the implied constant will not be $1$ (and more likely than not must depend upon $\epsilon$).

One such example which I can think of is that Mertens conjecture was proven false. This states that Mertens function which is given by $$ M(x)=\sum_{n\leq x}\mu(n) $$ satisfies $\vert M(x)\vert\leq \sqrt{x}$. I should note that $\mu(n)$ is the Möbius function in number theory which is $0$ unless $n$ is square-free in that case it is given by $$ \mu(p_1p_2...p_k)=(-1)^k $$ One reason that the disproof of this conjecture is disappointing is that Mertens conjecture implies the Riemann hypothesis, RH is equivalent to the statement that $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$. Thus, we expect the square root cancellation in the summation; however, the disappointing part of Mertens’ conjecture being false is that as much as we expect RH to be true, if one were to attempt to prove RH by showing $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$, then the implied constant will not be $1$ (and more likely than not must depend upon $\epsilon$).

One such example which I can think of is that Mertens conjecture was proven false. This states that Mertens function which is given by $$ M(x)=\sum_{n\leq x}\mu(n) $$ satisfies $\vert M(x)\vert\leq \sqrt{x}$. I should note that $\mu(n)$ is the Mobius function in number theory which is $0$ unless $n$ is square-free in that case it is given by $$ \mu(p_1p_2...p_k)=(-1)^k $$ One reason that the disproof of this conjecture is disappointing is that Mertens conjecture implies the Riemann hypothesis, RH is equivalent to the statement that $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$. Thus, we expect the square root cancelation in the summation; however, the disappointing part of Merten's conjecture being false is that as much as we expect RH to be true, if one were to attempt to prove RH by showing $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$, then the implied constant will not be $1$ (and more likely than not must depend upon $\epsilon$).

One such example which I can think of is that Mertens conjecture was proven false. This states that Mertens function which is given by $$ M(x)=\sum_{n\leq x}\mu(n) $$ satisfies $\vert M(x)\vert\leq \sqrt{x}$. I should note that $\mu(n)$ is the Mobius function in number theory which is $0$ unless $n$ is square-free in that case it is given by $$ \mu(p_1p_2...p_k)=(-1)^k $$ One reason that the disproof of this conjecture is disappointing is that Mertens conjecture implies the Riemann hypothesis, RH is equivalent to the statement that $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$. Thus, we expect the square root cancelation in the summation; however, the disappointing part of Mertens’ conjecture being false is that as much as we expect RH to be true, if one were to attempt to prove RH by showing $M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$, then the implied constant will not be $1$ (and more likely than not must depend upon $\epsilon$).