Conditional convergence of a series involving the Möbius function

The statement is appears to be misstated as written because proving that it converges depends on facts about the Riemann zeta function that are not proven. It is easy to prove that the series is not absolutely convergent, but the convergence of the signed series appears to be tied to the zeros of the Riemann zeta function.

For quadratfrei $b$, write $b=cd$. Then

$$ \frac1{b^2}\left(\frac1{c^2}+\frac{c^2}{b^2}\right)^{-3/2} =\frac{cd}{(c^2+d^2)^{3/2}}. $$

To see that it is not absolutely convergent, take $b=n(n+1)$ with both $n$ and $n+1$ quadratfrei. Then $b$ is quadratfrei, and the divisor $c=n$ occurs in the inner sum. The contribution is

$$ \frac{n(n+1)}{\bigl(n^2+(n+1)^2\bigr)^{3/2}}\asymp \frac1n. $$

A positive proportion of integers $n$ have both $n$ and $n+1$ quadratfrei. Therefore these terms alone yield a divergent harmonic-type sum, so

$$ \sum_b |\text{the $b$-th term}|=\infty. $$

The hard part is proving that the signed series converges. After symmetrizing the divisor sum and taking its Dirichlet series, you get an expression of the form

$$ D(z)=\frac1{\zeta(1+2z)} \int_{\bf R} \frac{\text{analytic factor}} {\zeta(\tfrac12+z+it/2)\zeta(\tfrac12+z-it/2)}\,dt. $$

An zero $\rho$ of $\zeta$ with $\Re \rho>\tfrac12$, would normally create a pole of $D(z)$ at

$$ z=\rho-\frac12, $$

which lies in the half-plane $\Re z>0$. But convergence of the original series would force $D(z)$ to be holomorphic there.

$$S(X)=\sum_{\substack{c<k,\ (c,k)=1\\ ck\le X}}\mu(c)\mu(k)\,\frac{ck}{(c^2+k^2)^{3/2}}$$

Consider primes $R<p<q\le 2R$. Then $b=pq$ is squarefree, $p<\sqrt{pq}$, and the absolute value of the corresponding term is at least $$K(p,q)=\frac{pq}{(p^2+q^2)^{3/2}}\ge \frac{R^2}{(8R^2)^{3/2}}=\frac1{16\sqrt2R}$$

\begin{multline*} \binom{\pi(2R)-\pi(R)}2\asymp \frac{R^2}{\log^2R}\Rightarrow \\\Rightarrow \sum_{b\le 4R^2}|a_b|\ge\frac1{16\sqrt2,R}\binom{\pi(2R)-\pi(R)}2\gg \frac{R}{\log^2R}\longrightarrow\infty \end{multline*}

Let $$E(x)=\exp\left[-c(\log x)^{3/5}(\log\log x)^{-1/5}\right],\quad M(x)\ll xE(x)$$

In the critical range $c,k\asymp R$, we have $K(c,k)\asymp R^{-1}$. Applying Abel summation twice together with the stated estimate yields, at best, a bound of order $\dfrac{M(R)^2}{R}\ll RE(R)^2$. However, $RE(R)^2\longrightarrow\infty$, so this estimate does not even show that the contribution of a single diagonal block tends to zero. By symmetry, set

$$q_b=\mu(b)\sum_{d\mid b}\frac{b}{\bigl(d^2+(b/d)^2\bigr)^{3/2}}\Rightarrow 2S(X)+2^{-3/2}=\sum_{b\le X}q_b$$ $$D(s)=\sum_{b\ge1}\frac{q_b}{b^s}=\sum_{m,n\ge1}\frac{\mu(mn)}{(mn)^{s+1/2}}h\left(\log\frac mn\right),\quad h(u)=(2\cosh u)^{-3/2},\quad \Re s>\frac12$$

After the Fourier transformations, we obtain the product $$\prod_p\left(1-p^{-s-1/2+it}-p^{-s-1/2-it}\right)=\frac{H(s,t)}{\zeta(s+\tfrac12-it),\zeta(s+\tfrac12+it),\zeta(1+2s)}$$

At $s=0$, the values $\zeta(\tfrac12\pm it)$ appear, i.e. precisely on the critical line of the zeta function. The zero of $1/\zeta(1+2s)$ may provide Abel regularization, but this does not imply ordinary convergence of the original partial sums