Can we find $\lim_{N\to \infty} \frac{1}{\sqrt{N}} \sum_{n \le N} \mu(n)/\sqrt{n}$

Question

Here $\mu(n)$ is möbius function.

Without assuming RH can we find if $\lim_{N \to \infty} \frac{1}{\sqrt{N}} \sum_{n=1}^N \frac{\mu(n)}{\sqrt{n}}$ exists, and if yes, what it may be. Calculations hint it may have a very small finite value or may go to zero.

One approach I can think of is using partial sum which goes to zero as $N \to \infty$ :

$$ \sum_{n \le N} \frac{\mu(n)}{n} = \sum_{n \le N} \frac{\mu(n)}{\sqrt{n}} \frac{1}{\sqrt{n}} $$ $$ = \frac{1}{\sqrt{N}} \sum_{n \le N} \frac{\mu(n)}{\sqrt{n}} - \int_1^N \frac{-1}{2t\sqrt{t}} \sum_{n \le t} \frac{\mu(n)}{\sqrt{n}} dt $$

Answer 1

Partial summation gives

$$ \frac{1}{\sqrt N} \sum_{n=1}^N \frac{\mu(n)}{\sqrt n} = \frac{M(N)}{N} + \frac{1}{\sqrt N} \sum_{n=1}^{N-1} M(n) \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)$$

where $ M(N) = \sum_{n=1}^N \mu(n) $ is the Mertens function. We know from the prime number theorem that $ M(x) = o(x) $, which gives the asymptotic

$$ = \frac{o(N)}{N} + \frac{1}{\sqrt N} \sum_{n=1}^{N-1} o(n^{-1/2}) = \frac{o(N)}{N} + N^{-1/2} o(\sqrt{N}) = o(1) \to 0 $$

No RH is needed for the result.