I want to estimate $\sum \limits_{n\leq x} \mu^2(n)$ with error term $o(\sqrt{x})$ Given that $\lim \limits_{x \to \infty} \frac{\psi(x)}{x}=1$ or that $\lim \limits_{x \to \infty} \frac{1}{x} \sum\limits_{n\leq x} \mu(x) = 0$
Without PNT i managed to estimate it with error term $O(\sqrt{x})$
My attempt : $\sum \limits_{n\leq x} \mu^2(n) = \sum \limits_{d^2 \leq x} \mu(d) [\frac{x}{d^2}] = x \sum \limits_{d \leq \sqrt{x}} \frac{\mu(d)}{d^2} + O( |\sum \limits_{n\leq \sqrt{x}} \mu(n)|) $ and since $M_{\mu} (\sqrt{x}) = o(\sqrt{x})$ it follows that $O( |\sum \limits_{n\leq \sqrt{x}} \mu(n)|)=o(\sqrt{x})$ ,but i don't know how to get error term $o(\sqrt{x})$ form $ x \sum \limits_{d \leq \sqrt{x}} \frac{\mu(d)}{d^2} = \frac{6x}{\pi^2} - x \sum \limits_{d > \sqrt{x}} \frac{\mu(d)}{d^2} $
Since $\bigl\lfloor \frac{x}{d^2}\bigr\rfloor = 0$ for $d > \sqrt{x}$ we can extend the sum giving $Q(x)$ to infinity, $$Q(x) = \sum_{d = 1}^{\infty} \mu(d)\biggl\lfloor \frac{x}{d^2}\biggr\rfloor = x\sum_{d = 1}^{\infty} \frac{\mu(d)}{d^2} - \sum_{d = 1}^{\infty} \mu(d) \biggl\lbrace \frac{x}{d^2}\biggr\rbrace\,,\tag{1}$$ where $\{u\} = u - \lfloor u\rfloor$ denotes the fractional part.
The first term on the right of $(1)$ is exactly $\frac{6}{\pi^2}\cdot x$, so what we need is to show $$\sum_{d = 1}^{\infty} \mu(d) \biggl\lbrace \frac{x}{d^2}\biggr\rbrace = o(\sqrt{x})\,.\tag{2}$$ Now the trick is to split the series at a suitable $y < \sqrt{x}$ (where $y$ depends on $x$ of course). For the first part we use the trivial estimate, $$\sum_{d = 1}^{y} \mu(d) \biggl\lbrace \frac{x}{d^2}\biggr\rbrace = O(y)$$ since $\lvert\mu(d)\rvert$ and fractional parts are bounded by $1$. Thus we must choose $y = o(\sqrt{x})$. The choice $y = \bigl\lfloor \sqrt{\frac{x}{m(x)}}\bigr\rfloor$ where the positive integer $m(x)$ tends to $\infty$ slowly as $x \to \infty$ is convenient.
To estimate the tail series, write $\mu(d) = M(d) - M(d-1)$ and apply summation by parts. If we have a strong enough bound for $M(x)$, we can see that $\sum \lvert M(d+c)\rvert\bigl\lbrace\frac{x}{d^2}\bigr\rbrace$ converges for every constant $c \geqslant 0$, so we can simply split and recombine the series. But with only $M(x) = o(x)$ available, a more cautious summation by parts is needed. For $z > y$ we have \begin{align} \sum_{d = y+1}^{z} \bigl(M(d) - M(d-1)\bigr)\biggl\lbrace \frac{x}{d^2}\biggr\rbrace &= M(z)\biggl\lbrace \frac{x}{(z+1)^2}\biggr\rbrace - M(y)\biggl\lbrace \frac{x}{(y+1)^2}\biggr\rbrace \\ &\qquad + \sum_{d = y+1}^{z} M(d)\biggl(\biggl\lbrace \frac{x}{d^2} \biggr\rbrace - \biggl\lbrace \frac{x}{(d+1)^2}\biggr\rbrace \biggr)\,. \end{align} The first term on the right tends to $0$ as $z \to \infty$. In the sum on the right hand side, $$\biggl\lbrace \frac{x}{d^2} \biggr\rbrace - \biggl\lbrace \frac{x}{(d+1)^2}\biggr\rbrace = \frac{x(2d+1)}{d^2(d+1)^2} - \underbrace{\biggl(\biggl\lfloor \frac{x}{d^2} \biggr\rfloor - \biggl\lfloor \frac{x}{(d+1)^2}\biggr\rfloor\biggr)}_{s(x,d)}\,,$$ where $s(x,d) = 0$, unless there is a positive integer $k$ such that $d = \bigl\lfloor \sqrt{\frac{x}{k}}\bigr\rfloor$, in which case it is $1$ if we choose $y$ not too small, $y > \sqrt[3]{3x}$, roughly equivalent to $m(x) < \sqrt[3]{\frac{x}{9}}$, suffices easily. Thus we can take the limit for $z \to \infty$ and obtain $$\sum_{d = y+1}^{\infty} \mu(d)\biggl\lbrace \frac{x}{d^2}\biggr\rbrace = -\sum_{k = 1}^{m(x)-1} M\biggl(\sqrt{\frac{x}{k}}\biggr) - M(y)\biggl\lbrace\frac{x}{(y+1)^2}\biggr\rbrace + \sum_{d = y+1}^{\infty} \frac{xM(d)(2d+1)}{d^2(d+1)^2}\,.$$ With $$\delta(u) := \sup \: \bigl\{ x^{-1}\lvert M(x)\rvert : x \geqslant u\bigr\}$$ we can then estimate $$\Biggl\lvert\sum_{k = 1}^{m(x)-1} M\biggl(\sqrt{\frac{x}{k}}\biggr)\Biggr\rvert \leqslant \delta(y)\sqrt{x}\sum_{k = 1}^{m(x)-1} \frac{1}{\sqrt{k}} \leqslant \delta(y)\sqrt{x}(2\sqrt{m(x)}+1)\,,$$ $\bigl\lvert M(y)\bigl\lbrace\frac{x}{(y+1)^2}\bigr\rbrace\bigr\rvert \leqslant y$ trivially, and $$\Biggl\lvert \sum_{d = y+1}^{\infty} \frac{xM(d)(2d+1)}{d^2(d+1)^2} \Biggr\rvert \leqslant 2\delta(y)x \sum_{d = y+1}^{\infty} \frac{1}{d(d+1)} < 2\delta(y)\sqrt{x}\cdot \sqrt{m(x)}\,.$$ Thus to prove the $o(\sqrt{x})$ error term we need to choose $m(x)$ (tending to $\infty$ with $x$) in such a way that $$\lim_{x \to \infty} \sqrt{m(x)}\delta\biggl(\biggl\lfloor \sqrt{\frac{x}{m(x)}}\biggr\rfloor\biggr) = 0\,.$$ We can do that: Choose a strictly increasing sequence $(\eta_k)$ in $[1,+\infty)$ such that $\delta(\eta_k) \leqslant \frac{1}{k}$. Put $\xi_k = k(\eta_k + 1)^2$ and $m(x) = k$ for $\xi_k \leqslant x < \xi_{k+1}$ (and $m(x) = 1$ for $x < \xi_1$). Clearly $m(x) \to \infty$ for $x\to \infty$, and for $\xi_k \leqslant x < \xi_{k+1}$ we have, using the monotonicity of $\delta$, \begin{align} \sqrt{m(x)}\delta\biggl(\biggl\lfloor \sqrt{\frac{x}{m(x)}}\biggr\rfloor\biggr) &= \sqrt{k}\delta\biggl(\biggl\lfloor \sqrt{\frac{x}{k}}\biggr\rfloor\biggr) \\ &\leqslant \sqrt{k}\delta\biggl(\sqrt{\frac{x}{k}}-1\biggr) \\ &\leqslant \sqrt{k} \delta\biggl(\sqrt{\frac{\xi_k}{k}} - 1\biggr) \\ &= \sqrt{k} \delta\bigl(\sqrt{(\eta_k+1)^2} - 1\bigr) \\ &= \sqrt{k}\delta(\eta_k) \\ &\leqslant \frac{1}{\sqrt{k}}\,. \end{align}
With explicit bounds for $\delta$ one can obtain explicit bounds for the error term. If we have a bound of the type $$\delta(x) = O\bigl(\exp\bigl(-c(\log x)^{\alpha}\bigr)\bigr)$$ with $c > 0$ and $0 <\alpha < 1$ — the exponent $\alpha = \frac{1}{2}$ corresponds to de la Vallée Poussin's zero-free region from 1899, the error term of Walfisz (not quite the same form, but similar enough) corresponds to that of Vinogradov/Korobov from 1958 — we can choose $$m(x) \approx \exp\bigl(b(\log x)^{\alpha}\bigr)$$ with an appropriate $0 < b < c$. Then $y = O\bigl(\sqrt{x}\,\exp(-b/2(\log x)^{\alpha})\bigr)$, and $$\log \sqrt{\frac{x}{m(x)}} \sim \frac{1}{2}\log x\biggl(1 - \frac{b}{(\log x)^{1-\alpha}}\biggr) \sim \frac{1}{2}\log x\,.$$ For every $0 < c' < c$ we therefore have $$\sqrt{m(x)}\delta\biggl(\biggl\lfloor \sqrt{\frac{x}{m(x)}}\biggr\rfloor\biggr) = O\biggl(\exp\biggl(-\biggl(\frac{c'}{2^{\alpha}} - \frac{b}{2}\biggr)(\log x)^{\alpha}\biggr)\biggr)\,.$$ The choice $b = c'\cdot 2^{-\alpha}$ yields $$Q(x) = \frac{6}{\pi^2}x + O\bigl(\sqrt{x}\,\exp\bigl(-b/2(\log x)^{\alpha}\bigr)\bigr)\,.$$
If we have a considerably stronger bound for $\delta$, say $\delta(x) = O(x^{-\beta})$ (meaning $\zeta(s)$ has no zeros with real part $> 1-\beta$), the loss in the strength of the error term becomes more obvious. We get the best order if we choose $m(x)$ such that $$\frac{1}{\sqrt{m(x)}} \asymp \sqrt{m(x)}\delta\biggl(\sqrt{\frac{x}{m(x)}}\biggr)$$ (that's also the case for bounds of the previous type), which occurs for $$m(x) \asymp x^{\beta/(2+\beta)}$$ and leads to an error bound of $O(x^{1/(2+\beta)})$ rather than the $\frac{1-\beta}{2}$ exponent one might hope for because $\zeta(s)/\zeta(2s)$ has no poles except at $1$ with real part $> \frac{1-\beta}{2}$.