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Here $\mu(n)$ is Möbius function and $M(x)$ is Mertens function.

The computations show that the partial sums $\sum_{n \le x} \frac{\mu(n)}{\sqrt{n}}$ stays between $-0.2$ and $-1.2$ when $10^1<x<10^8$, and it has been shown here that the series diverges.

I understand that Gonek's conjecture about how $M(x)/\sqrt{x}$ grows. The computations show that $M(x)/\sqrt{x}$ stays between $-0.5$ and $+0.5$ in the range $10^3<x<10^8$. Mertens - now disproved - conjecture, stated that $M(x)/\sqrt{x}$ would stay between $-1$ and $+1$, but for a very large number - may be as high as $~10^{10^{39}}$ - the bounds would be broken.

Is there a conjecture about the upper and lower bounds enveloping the partial sums $\sum_{n \le x} \frac{\mu(n)}{\sqrt{n}}$? (Here the bounds may be in form of function of x or constants).

RANGE for $\sum_{n<x} \frac{\mu(n)}{\sqrt{n}}$

1 to 10 |-0.7316706458761313 to 1.0

11 to 100 |-1.039366230708515 to -0.2773363002134797

101 to 1000 |-1.1146538644748882 to -0.2191909291869089

1001 to 10000 |-1.1181250910673983 to -0.2520110583510755

10001 to 100000 |-1.1381551083730033 to -0.24032822734981085

100001 to 1000000 |-1.1157717132476146 to -0.241077634847474

1000001 to 10000000 |-1.0814451758700396 to -0.2643931813309547

10000001 to 100000000 |-1.147086934330654 to -0.2772958213770883

RANGE for $\frac{M(x)}{\sqrt{x}}$

1 to 10 |-0.8944271909999159 to 1.0

11 to 100 | -0.8320502943378437 to 0.20519567041703082

101 to 1000 | -0.5671049640066687 to 0.3363363969981562

1001 to 10000 | -0.47220269325540665 to 0.4006822709402389

10001 to 100000 |-0.46297703636370996 to 0.4362149520604853

100001 to 1000000 |-0.43257122707105117 to 0.43777620594858513

1000001 to 10000000 |-0.40430689124572694 to 0.41824547582196664

10000001 to 100000000 |-0.4627286901166685 to 0.41265902449721137

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    $\begingroup$ This is really the same as the Mertens conjecture in the form $M(x)=O(x^{1/2})$. Wikipedia says that there is a conjecture that it is false with $M(x)\approx O(x^{1/2}(\log\log \log x)^a) $ for some $a>0$. $\endgroup$
    – reuns
    Commented Mar 17, 2021 at 19:56
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    $\begingroup$ As @reuns mentions, it should be the same as for the Mertens function except divided through by $\sqrt{x}$. In particular, the natural conjecture is that $$0 < \limsup_{x \to \infty} (\log \log \log x)^{-5/4} \sum_{n \leq x} \mu(n) n^{-1/2} < \infty$$ and $$-\infty < \liminf_{x \to \infty} (\log \log \log x)^{-5/4} \sum_{n \leq x} \mu(n) n^{-1/2} < 0.$$ See also my answer here: mathoverflow.net/a/368511/3803 $\endgroup$
    – Peter Humphries
    Commented Mar 17, 2021 at 20:04
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    $\begingroup$ Also, there is a bias in this partial sum; assuming standard conjectures, $\sum_{n \leq x} \mu(n) n^{-1/2}$ should have a limiting logarithmic distribution whose median and mean is equal to $1/\zeta(1/2) \approx -0.684765\ldots$. Nonetheless, it still changes sign infinitely often. $\endgroup$
    – Peter Humphries
    Commented Mar 17, 2021 at 20:06
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    $\begingroup$ If $\sum_{n\le x} a_n$ doesn't change of sign infinitely often then $F(s)=\int_1^\infty (\sum_{n\le x}a_n)x^{-s-1}dx$ has a singularity at $\sigma$ its abscissa of convergence. $a(n)=\mu(n)/n$ gives $F(s)=1/(s\zeta(s+1))$, no singularity on $[-1/2,0]$ whence $\sum_{n\le x} \mu(n)/n$ changes of sign infinitely often. $1/\zeta(1/2)-\sum_{n\le x} \mu(n)n^{-1/2}$ gives $F(s)=\frac1{s\zeta(1/2+s)}-\frac1{s\zeta(1/2)}$, no singularity on $[0,1/2]$ whence $1/\zeta(1/2)-\sum_{n\le x} \mu(n) n^{-1/2}$ changes of sign infinitely often. $\endgroup$
    – reuns
    Commented Mar 18, 2021 at 0:14
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    $\begingroup$ For $c\ne 0$, $\int_1^\infty (M(x)-cx^{1/2})x^{-s-1}dx= \frac1{s\zeta(s)}-\frac1{c (s-1/2)}$ does have a singularity at $1/2$ so under the RH we can't tell if $M(x)-cx^{1/2}$ changes of sign infinitely often. $\endgroup$
    – reuns
    Commented Mar 18, 2021 at 0:20

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