Put
$$
A(x)=\sum_{p\leq x}\frac{\log p}{p}=\log x+E(x),
\qquad E(x)=O(1).
$$
The <Abel summation formula> with weight $1/\log t$ gives
$$
\sum_{p\leq x}\frac1p
=\frac{A(x)}{\log x}
+\int_2^x\frac{A(t)}{t(\log t)^2}\,dt.
$$
Substituting $A(t)=\log t+E(t)$ yields
$$
\sum_{p\leq x}\frac1p
=\log\log x+c+\frac{E(x)}{\log x}
-\int_x^\infty\frac{E(t)}{t(\log t)^2}\,dt
$$
for a constant $c$. Both final terms are $O(1/\log x)$, so
$$
\boxed{\sum_{p\leq x}\frac1p
=\log\log x+c+O\left(\frac1{\log x}\right).}
$$
This is the <Mertens second theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / b / Solution

The same divisor-identity argument as in part (a) gives
$$
\sum_{n\leq x}\frac{\Lambda(n)}n=\log x+O(1).
$$
On the other hand, the <Abel summation formula> gives
$$
\sum_{n\leq x}\frac{\Lambda(n)}n
=\frac{\psi(x)}x+\int_1^x\frac{\psi(t)}{t^2}\,dt.
$$
Under the proposed asymptotic, the right side is
$$
a\log x+b\log\log x+o(\log\log x)+O(1).
$$
Dividing first by $\log x$ gives $a=1$. Subtracting $\log x$, dividing by $\log\log x$, and taking the <limit> then gives $b=0$. Therefore
$$
\boxed{a=1,\qquad b=0.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / c / Solution

For $\sigma>1$, <Orthogonality of Dirichlet characters> gives
$$
\sum_{\substack{n\geq1\\n\equiv a\pmod q}}
\frac{\Lambda(n)}{n^\sigma}
=\frac1{\varphi(q)}
\sum_{\chi\bmod q}\overline{\chi(a)}
\left(-\frac{L'(\sigma,\chi)}{L(\sigma,\chi)}\right).
$$
The principal-character term is
$$
\frac1{\sigma-1}+O_q(1).
$$
Part (a) makes every nonprincipal logarithmic derivative bounded as $\sigma\to1^+$, so
$$
\sum_{\substack{n\geq1\\n\equiv a\pmod q}}
\frac{\Lambda(n)}{n^\sigma}
=\frac1{\varphi(q)(\sigma-1)}+O_q(1)
\longrightarrow\infty.
$$
If the nondecreasing <Chebyshev function in an arithmetic progression>
$$
\psi(x;q,a)=\sum_{\substack{n\leq x\\n\equiv a\pmod q}}\Lambda(n)
$$
were bounded, the <Abel summation formula> would keep the displayed Dirichlet series bounded near $\sigma=1$. Therefore
$$
\boxed{\psi(x;q,a)\longrightarrow\infty.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 4 / b / Solution

{c}
{wiki=Abel's_summation_formula}

= Partial summation
{synonym}

= Summation by parts
{synonym}

Partial summation is the discrete analogue of <integration by parts>. If $A(x)=\sum_{n\leq x}a_n$ and $g$ is continuously differentiable, then
$$
\sum_{n\leq x}a_ng(n)=A(x)g(x)-\int_1^xA(t)g'(t)\,dt.
$$

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