# Riemann Hypothesis and $\sum\limits_{k\leq n}\left(\frac{\mu(k)}{k}\right)^2$

I know that Riemann Hypothesis is equivalent to the following statement

$\sum\limits_{k\leq n}\frac{\mu(k)}{k}=O(n^{-1/2+\epsilon})$

Is there any relation between Riemann Hypothesis and $\sum\limits_{k\leq n}\left(\frac{\mu(k)}{k}\right)^2$

Your statement is provably false (consider $n$ prime).

The correct statement is that RH is equivalent to $\sum_{k\le n} \frac{\mu(k)}k = O(n^{-1/2+\varepsilon})$. This is related to the fact that $\sum_{k=1}^\infty \frac{\mu(k)}{k^s} = \frac1{\zeta(s)}$ where the series converges, and in particular, $\sum_{k=1}^\infty \frac{\mu(k)}k = 0$. (Indeed, the statement talks about the rate of convergence of the full series $\sum_{k=1}^\infty \frac{\mu(k)}k$ to its limit.)

On the other hand, $\sum_{k\le n} \frac{\mu^2(k)}{k^2}$ is related to the series $\sum_{k=1}^\infty \frac{\mu^2(k)}{k^s} = \frac{\zeta(s)}{\zeta(2s)}$, for which $\sum_{k=1}^\infty \frac{\mu^2(k)}{k^2} = \frac{\zeta(2)}{\zeta(4)} = \frac{15}{\pi^2}$. However, the rate of convergence of the series $\sum_{k=1}^\infty \frac{\mu^2(k)}{k^2}$ to its limit isn't related to the zeros of $\zeta(s)$. Indeed, this series has all nonnegative terms, so it's easy to see that $\sum_{k\le n} \frac{\mu^2(k)}{k^2} - \frac{15}{\pi^2} = O(n^{-1})$ and that this error term is best possible. (I don't immediately know what we know about the error term in the better approximation $\sum_{k\le n} \frac{\mu^2(k)}{k^2} - \frac{15}{\pi^2} + \frac6{\pi^2n}$.)

• thank you so much for correcting my problem. Commented Jul 28, 2016 at 14:08