I have given the following lemma:
Let $\varepsilon>0$ and let $a, q, z$ be such that $$ 1 \leqslant a \leqslant q \leqslant B^{2}, \quad \operatorname{gcd}(a, q)=1, \quad|z| \leqslant \frac{1}{q^{2}} $$ Then we have $$ S\left(\frac{a}{q}+z\right) \ll_{\varepsilon} B^{n+\varepsilon} q^{-\frac{n}{24}} \min \left\{1,|z| B^{4}\right\}^{-\frac{n}{24}} $$
Furthermore the major arcs are defined as follows:
$\mathfrak{M}_{q, a}(\Delta)=\left[\frac{a}{q}-B^{-4+\Delta}, \frac{a}{q}+B^{-4+\Delta}\right]$
for $1 \leqslant a \leqslant q$ with $(a, q)=1$ and $q \leqslant B^{\Delta} .$
With $\Delta$ being a small positive parameter.
The full set of major arcs is given by \begin{equation} \mathfrak{M}(\Delta)=\bigcup_{1 \leqslant q \leqslant B^{\Delta}} \bigcup_{1 \leqslant a \leqslant q \atop \operatorname{gcd}(a, q)=1} \mathfrak{M}_{q, a}(\Delta) \end{equation}
We define the minor arcs modulo 1 as: $\mathfrak{m}(\Delta)=[0,1] \backslash \mathfrak{M}(\Delta)$.
Now I want to estimate the minor arcs. So let $\alpha \in \mathfrak{m}(\Delta)$. By Dirichlet's approximation theorem we can find integers $a$ and $q$ such that $(a,q)=1$ and $1 \leq a \leq q \leq B^2$. Therefore $|\alpha-\frac{a}{q}| \leq \frac{1}{B^2q}$. From the definition of the major arcs we see that for $\alpha \in \mathfrak{m}(\Delta) = [0,1] \setminus \mathfrak{M}(\Delta)$ we may assume that $q > B^{\Delta}$ or $|q\alpha - a| > qB^{-4+\Delta}$.
Writing $\alpha=\frac{a}{q}+z$ we can now apparently conclude from the lemma that: $$ \begin{gathered} \int_{\mathfrak{m}(\Delta)}|S(\alpha)| \mathrm{d} \alpha \ll_{\varepsilon} B^{n+\varepsilon} \sum_{q \leqslant B^{2}} q^{1-\frac{n}{24}} \int_{\frac{-1}{q B^{2}}}^{\frac{1}{q B^{2}}} \min \left\{1,|z| B^{4}\right\}^{-\frac{n}{24}} \mathrm{~d} z \\ \ll_{\varepsilon} B^{n-4-\Delta\left(\frac{n}{24}-2\right)+\varepsilon} \end{gathered} $$ under the assumption that $n>48$.
I sadly don't really understand the last two estimations. I don't really see, where this $\sum_{q \leqslant B^{2}} q^{1-\frac{n}{24}}$ is coming from. And in the last estimation I'm kind of lost as well. Any hint or help would be much appreciated. :)