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I’ve asked this, after I was doing the calculations that I show below

Question.Let $\mu(n)$ the Möbius function, and $\pi(n)$ the prime-counting function. How one can to prove that $$\sum_{n=1}^\infty\frac{\mu(n)\pi(n)}{n^2}$$ does converge?

The calculations of my motivation were to combine the Prime Number Theorem and this asymptotic formula 27.12.3 from the Digital Library of Mahtematical Functions, to calculate $$\sum_{n=1}^\infty\frac{\mu(n)\pi(n)}{n^2}=0-\frac{6}{\pi^2}-\sum_{n=1}^\infty\frac{\mu(n)}{n^2}\sum_{p_k\leq \sqrt{n}}\lfloor\frac{n}{p_k}\rfloor$$

$$\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{n=1}^\infty\frac{\mu(n)}{n^2}\left(\sum_{r\geq 2}(-1)^r\sum_{p_{k_1}<\cdots<p_{k_r}\leq \sqrt{n}}\lfloor\frac{n}{p_{k_1}\cdots p_{k_r} }\rfloor\right).$$

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Can you analyse the summands in previous RHS, or is better deduce that $\sum_{n=1}^\infty\frac{\mu(n)\pi(n)}{n^2}$ does converge by summation or a different method? If one can deduce using summation only are required hints.

Wolfram Alpha tell us that seems that is convergent $\approx -0.60$

sum mu(n)PrimePi(n)/n^2, from n=1 to 10000

**Thanks in advance.**

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Assuming we already know that

$$\sum_{n = 1}^\infty \frac{\mu(n)}{n}$$

converges (to $0$, but that’s irrelevant), for $n \geqslant 2$ write

$$\frac{\pi(n)}{n} = \frac{1}{\log n} + \biggl(\frac{\pi(n)}{n} – \frac{1}{\log n}\biggr).$$

By the prime number theorem, $\frac{\pi(n)}{n} – \frac{1}{\log n} \in O\bigl((\log n)^{-2}\bigr)$, so

$$\sum_{n = 2}^\infty \frac{\mu(n)}{n}\biggl(\frac{\pi(n)}{n} – \frac{1}{\log n}\biggr)$$

converges absolutely. The sequence $\frac{1}{\log n}$ converges to $0$ and is monotonically decreasing, so by Dirchlet’s test

$$\sum_{n = 2}^\infty \frac{\mu(n)}{n}\cdot \frac{1}{\log n}$$

converges (conditionally). It follows that

$$\sum_{n = 1}^\infty \frac{\mu(n)\pi(n)}{n^2} = \sum_{n = 2}^\infty \frac{\mu(n)\pi(n)}{n^2}$$

converges (conditionally), as it is the sum of two convergent series.

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