We set, for all real $t \geqslant 2$,
$$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$
Show, using the result from question 16, that
$$\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} = 1 + \ln_2(n) - \ln_2(2) + \frac{R(n)}{\ln(n)} + \int_2^n \frac{R(t)}{t(\ln(t))^2} \, \mathrm{d}t$$