Studying and working with some problems involving the Möbius function, I (erroneously and somewhat randomly) found the following series/limit. I am curious about it, since I don't know wether it converges or not.
$$\lim_{x \to \infty} \sum_{1 \le n < x} \mu(n) \left \lfloor \frac{x}{n} \right \rfloor \log \left ( 1-\frac{n}{x} \right )$$
Since
$$\sum_{1 \le n < x} \log \left ( 1-\frac{n}{x} \right )=O(x)$$
Or
$$O\left ( \sum_{1 \le n \le x} 1 \right )$$
And by Generalised Möbius Inversion we know that
$$1= \sum_{1 \le n \le x} \mu(n) \left \lfloor \frac{x}{n} \right \rfloor$$
It seems reasonable for the limit to converge to a constant.
However, we are not allowed to use traditional methods to work with it for a lot of reasons (no integral tests, no summation by parts, etc.).
Any hint towards the solution of the problem or advice about how to approach to this kind of series and limits would be useful. Thank you very much.
If we set $$ f(x) = \sum_{1 \le n < x} \mu(n) \left \lfloor \frac{x}{n} \right \rfloor \log \left ( 1-\frac{n}{x} \right ), $$ then $\lim_{x\to\infty} f(x)$ cannot exist. Note that when $p$ is prime, none of the individual summands changes when $x$ goes from just below $p$ to just above $p$; therefore the change in $f(x)$ near $x=p$ is only from the new term $\mu(p) \left \lfloor \frac{x}{p} \right \rfloor \log \left ( 1-\frac{p}{x} \right ) = -\log \left ( 1-\frac{p}{x} \right )$. This function is unbounded for $x$ near $p$, so $f(x)$ is not even bounded, let alone convergent as $x\to\infty$.
Even if we restrict $x$ to integers, setting $x=p+1$ results in a new term of size $\log p$ or so, which is still unbounded. (One then has to deal with the terms where $n$ divides $p+1$, but the total change in those terms is $\ll \tau(p+1)$, the number of divisors of $p+1$, and that is $o(\log p)$.)