For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Let $\left(a_n\right)_{n \geqslant 2}$ be a sequence of real numbers. For $t \in \mathbb{R}$, we set $A(t) = \sum_{2 \leqslant k \leqslant t} a_k$. Let $b : [2, +\infty[ \rightarrow \mathbb{R}$ be a function of class $\mathscr{C}^1$. Show that for any integer $n \geqslant 2$, $$\sum_{k=2}^{n} a_k b(k) = A(n)b(n) - \int_2^n b'(t) A(t) \, \mathrm{d}t$$
For any non-zero natural integer $n$, we set
$$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$
Let $\left(a_n\right)_{n \geqslant 2}$ be a sequence of real numbers. For $t \in \mathbb{R}$, we set $A(t) = \sum_{2 \leqslant k \leqslant t} a_k$. Let $b : [2, +\infty[ \rightarrow \mathbb{R}$ be a function of class $\mathscr{C}^1$. Show that for any integer $n \geqslant 2$,
$$\sum_{k=2}^{n} a_k b(k) = A(n)b(n) - \int_2^n b'(t) A(t) \, \mathrm{d}t$$