# Non-negativity of an infinite absolutely convergent sum

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.

Thanks in advance for any help.

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 $$2. If such a $$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, 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 $$1/2$$.