Large gap between two consecutive square-free numbers Let $q_n$ denote the $n$-th square-free number. By Chinese remainder theorem (see this post), it is not difficult to show that there is arbitrarily large gap between two consecutive square-free numbers, i.e., $\limsup_{n\to\infty}(q_{n+1}-q_n)=\infty$. How to prove the stronger bound?
$$\limsup_{n\to\infty}\frac{q_{n+1}-q_n}{\log n/\log\log n}\geq \frac{1}{2}.$$
I don't how to get started, thanks for any help.
Edit: This is an exercise (Exercise 2.20) from A.J. Hildebrand's An introduction to analytic number theory.
 A: 
How to prove the stronger bound?

By estimating the index where the constructed gap between successive squarefree numbers occurs. Since the denominator, $\frac{\log n}{\log \log n}$ increases with $n$ for $n > e^e$, overestimating the index underestimates the quotient.
Thus, if to construct a gap of size $\geqslant g$ we find $m$ satisfying the congruences
$$m \equiv -k \pmod{p_k^2}$$
for $1 \leqslant k < g$, by the Chinese remainder theorem there is a unique solution satisfying
$$0 < m < (p_{g-1}\#)^2$$
(where $x\#$ is the primorial of $x$, i.e. the product of all primes not exceeding $x$). Then we have a gap $\geqslant g$ between $q_n$ and $q_{n+1}$, where $n = Q(m)$ is the number of squarefree numbers $\leqslant m$.
We know that $Q(m) = \frac{6}{\pi^2}m + O(\sqrt{m})$, but that doesn't give anything beyond the trivial $n = Q(m) \leqslant m$. We must, however, assume $g \geqslant 4$ to ensure $n > e^e$, so that our overestimating of $n$ certainly doesn't increase the quotient. The bound $n < (p_{g-1}\#)^2 < (p_g\#)^2$ yields
$$\log n < 2 \log (p_g\#) = 2 \vartheta(p_g)$$
and hence
\begin{align}
\frac{q_{n+1} - q_n}{\frac{\log n}{\log\log n}} &\geqslant \frac{g}{\frac{\log n}{\log\log n}} \\
&> \frac{g}{\frac{2\vartheta(p_g)}{\log \vartheta(p_g) + \log 2}} \\
&\sim \frac{g}{\frac{2p_g}{\log p_g}} \\
&\sim \frac{1}{2}
\end{align}
using the prime number theorem for $\vartheta(p_g) \sim p_g \sim g\log g$ and $\log \vartheta(p_g) = \log p_g + O(1) \sim \log g$.
For the following sharper estimate
$$\limsup_{n \to \infty} \frac{q_{n+1} - q_n}{\frac{\log n}{\log \log n}} \geqslant \frac{\pi^2}{12}$$
one uses the fact that each of the small primes produces not only one non-squarefree number in the gap but approximately $g/p^2$. Thus let $c > 1$ arbitrary, and choose $r$ such that
$$\prod_{\rho = 1}^{r} \biggl(1 - \frac{1}{p_{\rho}^2}\biggr) < c\cdot \frac{6}{\pi^2}\,.$$
Then, for sufficiently large $g$ every sequence of $g$ consecutive integers contains fewer than
$$c^2\frac{6}{\pi^2}\cdot g$$
integers that aren't divisible by any of the $p_{\rho}^2$, $1 \leqslant \rho \leqslant r$.
Hence to produce a gap of size at least $g$ (for sufficiently large $g$) we don't need $g-1$ primes,
$$A(g) = r + \biggl\lfloor c^2\frac{6}{\pi^2}g\biggr\rfloor$$
primes suffice. Then in the computation above we can replace $p_g$ with $p_{A(g)}$, yielding
$$\frac{q_{n+1}-q_n}{\frac{\log n}{\log \log n}} > \frac{g}{2\frac{\vartheta(p_{A(g)})}{\log \vartheta(p_{A(g)})}} \sim \frac{g}{2 A(g)} \sim \frac{\pi^2}{12c^2}\,.$$
Since $c > 1$ was arbitrary,
$$\limsup_{n \to \infty} \frac{q_{n+1}-q_n}{\frac{\log n}{\log \log n}} \geqslant \frac{\pi^2}{12}\,.$$
