Abel's summation formula

Partial summation is the discrete analogue of integration by parts. If $A(x)={\sum }_{n\le x}{a}_n$ and $g$ is continuously differentiable, then

If $A(x)$ counts a set of positive integers and $\eta >0$, Abel summation gives ${\sum }_{n\in A,n\le X}1/n=A(X)/X+{\int }_1^{X}A(t)t^{-2}dt$. A bound $A(t)\ll t/(\log t{)}^{1+\eta }$ for $t\ge 3$ makes the integral convergent, since the substitution $u=\log t$ gives ${\int }_{\log 3}^{\infty }{u}^{-1-\eta }du<\infty$.

If $\sum a_n$ converges and $b_n$ is bounded with $\sum |b_{n+1}-b_n|<\infty$, then $\sum a_n{b}_n$ converges. For $T_j={\sum }_{k=m}^j{a}_k$, summation by parts givesThe first factor tends uniformly to zero by the Cauchy sequence criterion, and the second has a uniform bound. A bounded monotone sequence has finite total variation, so monotone multipliers are included. Eventual monotonicity also suffices, since finitely many initial terms do not affect convergence. Absolute convergence of $\sum a_n$ is unnecessary.

For $3\le z\le D$, prime-number estimates and partial summation give, for an absolute $c>0$,Indeed, comparison with the prime-density integral and the substitution $v=\log D/\log t$ reduce the left side to $(\log D{)}^{-3}{\int }_{\log D/\log z}^{\infty }{v}^2{e}^{-v/2}dv$.