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.
Best Answer
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 $1/2$.