Let $\mu(n)$ the Möbius function. I know that combining Abel summation formula, the Prime Number Theorem and l'Hôpital's rule I can deduce $$\lim_{x\to\infty}\frac{1}{x}\sum_{2\leq n\leq x}\mu(n)\cdot\frac{\log\log n}{{\log\log (n+1)}}= 0-\lim_{x\to\infty}\frac{1}{x}\int_2^x M(t)\cdot f'(t)dt,$$ where $M(x)$ is the Mertens function, and $f(x)=\frac{\log\log x}{{\log\log (x+1)}}$. I know that $M(t)$ can be bounded by the obvious $t$, thus RHS should be
$$\lim_{x\to\infty}\frac{1}{x}O\left(\int_2^x \frac{(t+1)\log(t+1)\cdot\log\log(t+1)-t\log t\cdot\log\log t)}{(t+1)\log t\cdot\log(t+1)\cdot\log^2\log(t+1)}dt\right).$$
Question. Is there a reasonable way to compute (or compute improving my bound for Mertens function without additional more tedious computations) the limit in RHS? I say, if it is possible, determine if there is convergence and, if there is a limit compute it without do the more hardest computations. Thanks in advance.
Now with the BOUNTY I am looking an approach that provide us the behaviour of the first series, thus RHS, thus if it neccesary tedious computations your answer will be welcome. Thanks.
Note that holds $$\frac{\log\left(\log\left(n\right)\right)}{\log\left(\log\left(n+1\right)\right)}\sim1$$ as $n\rightarrow\infty$ so your sum is very close to $M\left(x\right)$. Then $$\frac{1}{x}\sum_{2\leq n\leq x}\mu\left(n\right)\frac{\log\left(\log\left(n\right)\right)}{\log\left(\log\left(n+1\right)\right)}\sim\frac{1}{x}\sum_{2\leq n\leq x}\mu\left(n\right)=\frac{1}{x}\left(M\left(x\right)-1\right)\rightarrow0$$ as $x\rightarrow\infty$ since PNT is equivalent to $$\frac{M\left(x\right)}{x}\stackrel{x\rightarrow\infty}{\longrightarrow}0.$$ Maybe it's interesting to claim a related result which can be found in the Terence Tao blog (with the references).