The infinite sums involving mobius function and a multiplicative function has got quite interest in past. In particular, sums of the form $$\sum\_{d=1}^{\infty}\frac{\mu(d)}{f(d)}$$ for mobius function $\mu$ and multiplicative function $f$ have been investigated for various $f.$ I am interested in knowing about any arguments that could prove/disprove the non-negativity of the following sum $$\sum\_{d=1}^{\infty}\frac{\mu(d)}{\mathrm{lcm}(d,\varphi(d))}$$ where $\varphi$ is the euler totient function. The function $f(d)=\mathrm{lcm}(d,\varphi(d))$ is not multiplicative and hence any standard techniques of treating multiplicative $f$ won't work here.


I would like to remark that the sum is absolutely convergent. To see this, one can consider the Lucas sequence $u\_n=2^n-1$ and let $\mathrm{ord}\_n(2)$ denote the multiplicative order of $2$ modulo $n.$ It is well known that $\mathrm{ord}\_n(2)\mid \varphi(n).$ This gives that $$\mathrm{lcm}(n,\varphi(n))\ge \mathrm{lcm}(n,\mathrm{ord}\_n(2)).$$ Thus, we have that $$\sum\_{d=1}^{\infty}\frac{1}{\mathrm{lcm}(d,\varphi(d))}\le \sum\_{d=1}^{\infty}\frac{1}{\mathrm{lcm}(d,\mathrm{ord}\_d(2))}$$ and the convergence of right sum follows by proposition 1.4 in [this published paper.](https://arxiv.org/abs/1805.02225)


Thanks in advance for any help.



Non-negativity of an infinite absolutely convergent sum
In general, it is better to approach such a question numerically, since your sum is absolutely convergent. However, in your particular case, it is possible to compute this explicitly without any numerical calculations. Notice that non-zero summands that appear in your sum correspond to squarefree $d$ (otherwise $\mu(d)=0$). Next, take a large $X$ and consider all squarefree $d$ with $2<d\leq X$. If such a $d$ is even, then $d=2d\_1$ and $d\_1>1$ is squarefree and odd. Since $\varphi(d)=\varphi(d\_1)$ is even, we have $[d,\varphi(d)]=[d\_1,\varphi(d\_1)]$ (here $[a,b]=\mathrm{lcm}(a,b)$). On the other hand, $\mu(d)=\mu(2d\_1)=-\mu(d\_1)$, therefore
$$
\frac{\mu(d)}{[d,\varphi(d)]}+\frac{\mu(d\_1)}{[d\_1,\varphi(d\_1)]}=0.
$$
Every even squarefree $d$ in $(2,X]$ is a member of one such pair, and same is true for odd squarefree $d$ with $1<d\leq X/2$, so
$$
\sum\_{d\leq X}\frac{\mu(d)}{[d,\varphi(d)])}=1-\frac12+\sum\_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}.
$$
This last summand can be estimated as follows
$$
\left|\sum\_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}\right|\leq \sum\_{\text{odd }d>X/2}\frac{1}{[d,\mathrm{ord}\_2(d)]}\ll \exp(-1/3(\ln X\ln\ln X)^{1/2})=o(1),
$$
by the paper you linked. Therefore, your sum is equal to $1/2$.


