Using a very complicated argument, I believe I can prove:
$$ 1 = M([p-1]) + M([\frac{p-1}{2}]) + M([\frac{p-1}{3}]) + \dots $$
Where $M(x)$ is Merten's function $p$ is any arbitrary prime and $[x]$ is the greatest integer function. I was wondering if there existed any alternative proofs?
Some remarks. Fix $n.$ Now notice that the term $\mu(k)$ appears in the sum $\sum\limits_{k=1}^{\lfloor(p-1)/n\rfloor}\mu(k)$ if and only if $n\leqslant\lfloor(p-1)/k\rfloor$ so the term $\mu(k)$ appears exactly $\lfloor(p-1)/k\rfloor$ times in $\sum\limits_{n=1}^\infty\sum\limits_{k=1}^{\lfloor(p-1)/n\rfloor}\mu(k)=\sum\limits_{n=1}^\infty M\left(\frac{p-1}{n}\right).$ Thus $\sum\limits_{n=1}^\infty M\left(\frac{p-1}{n}\right)=\sum\limits_{n=1}^\infty\mu(n)\lfloor(p-1)/n\rfloor=\sum\limits_{n=1}^{p-1}\mu(n)\lfloor(p-1)/n\rfloor.$
Also notice that, for each $1\leqslant n\leqslant p-1,$ the term $\mu(n)$ appears exactly $\lfloor(p-1)/n\rfloor$ times in $\sum\limits_{n=1}^{p-1}\sum\limits_{d\mid n}\mu(d)$ because there are exactly $\lfloor(p-1)/n\rfloor$ multiples of $n$ between $1$ and $p-1$ so we have $\sum\limits_{n=1}^{p-1}\sum\limits_{d\mid n}\mu(d)=\sum\limits_{n=1}^{p-1}\mu(n)\lfloor(p-1)/n\rfloor$ and since $\sum\limits_{d\mid n}\mu(d)=1$ if $n=1$ and $\sum\limits_{d\mid n}\mu(d)=0$ if $n>1$ then $1=\sum\limits_{n=1}^{p-1}\sum\limits_{d\mid n}\mu(d)=\sum\limits_{n=1}^{p-1}\mu(n)\lfloor(p-1)/n\rfloor.$