I've been reading through Atle Selberg's "Elementary" proof of the Prime Number Theorem. That seems to be an ironic usage of the word. A lot of discussions focus on the Selberg formula:
$$ \theta(x) \;\log x + \sum_{p \leq x} \log p \; \theta\big(\frac{x}{p}\big) = 2x \log x + O(x) $$
The left side is almost suggesting to let $p=1$ be a prime number. Here $\theta$ is the sum of logs of primes:
$$ \theta(x) = \sum_{p \leq x} \log p = \log \bigg[ \prod_{p \leq x} p \bigg]$$ Here emphasizing the product of all prime numbers (rather than a factorial), which can be used to prove the infinitude of primes. We can write as correlations of the sequence $\{ \log p: p \text{ prime}\}$.
$$ \sum_{pq \leq x} \log p \, \log q + \sum_{p \leq x} \log^2 p = 2x \log x + O(x) $$
Let me go back a few steps and go through three steps in Selberg's derivation.
$$ \sum_{p \leq x} \log p = x\log x + O(x) \tag{A} $$
This step easily implies
$$ \sum_{pq \leq x} \frac{\log p \log q}{pq} = \frac{1}{2} \log^2 x + O(\log x) \tag{B}$$
and then we can do a partial summation to get :
$$ \sum_{pq \leq x} \frac{\log p \log q}{pq (\log p + \log q)} = \log x + O(\log \log x) \tag{C}$$
In fact I do not see at all how (A) → (B) → (C)
I simply do not have the facility to do all the re-arrangments he talks about. Here is summation by parts:
$$ \sum_{k=m}^n f_k \Delta g_k = [f_{n+1}g_{n+1}-f_mg_m]-\sum_{k=m}^n g_{k+1}f_k $$
Summation by parts is not the same as partial summation, usually he refers to Abel's summation formula, namely that $$\sum_{y<n\leq x}a(n)f(n)=A(x)f(x)-A(y)f(y)-\int_y^xA(t)f'(t)dt$$ where $f$ is differentiable on $[y,x]$ and $A(x)=\sum_{n\leq x}a(n)$ with $A(x)=0$ for $x<1$.