Simplest proof that $\sum 1/p$ grows like $\log \log x$ I am seeking recommendations for a simple proof that $\sum_{p<x} 1/p$ grows like $\log (\log (x))$.
By simple I mean suitable for secondary school or 1st year university students - not those who are already trained in mathematics.
Ideally the proof would start from Euler's product formula for the Riemann zeta function, but this is not necessary.
 A: Without the knowledge of partial summation, we just use the traditional summation by parts:
Let $\pi(x)$ denote the number of prime numbers less than or equal to $x$, so for all $n\in\mathbb Z^{>0}$
$$
\pi(n)-\pi(n-1)=
\begin{cases}
1 & n\text{ is prime} \\
0 & \text{otherwise}
\end{cases}
$$
Applying this to our problem, we have
$$
\begin{aligned}
\sum_{p\le N}\frac1p
&=\sum_{n=1}^N{\pi(n)-\pi(n-1)\over n}={\pi(N)\over N}+\sum_{n=1}^{N-1}{\pi(n)\over n}-\sum_{n=1}^N{\pi(n-1)\over n} \\
&={\pi(N)\over N}+\sum_{n=2}^{N-1}{\pi(n)\over n}-\sum_{n=2}^{N-1}{\pi(n)\over n+1}={\pi(N)\over N}-\sum_{n=2}^{N-1}\pi(n)\left[{1\over n+1}-\frac1n\right] \\
&={\pi(N)\over N}+\sum_{n=2}^{N-1}\pi(n)\int_n^{n+1}{\mathrm dt\over t^2}={\pi(N)\over N}+\int_2^N{\pi(t)\over t^2}\mathrm dt
\end{aligned}
$$
Let $N=\lfloor x\rfloor$, then
$$
\int_N^x{\pi(t)\over t^2}\mathrm dt=\pi(N)\int_N^x{\mathrm dt\over t^2}={\pi(N)\over N}-{\pi(x)\over x}
$$
As a result, the above formula applies to all $x\in\mathbb R_{>0}$:
$$
\sum_{p\le x}\frac1p={\pi(x)\over x}+\int_2^x{\pi(t)\over t^2}\mathrm dt
$$
By the prime number theorem, we know that there exists a positive constant $K$ such that
$$
\left|\pi(x)-{x\over\log x}\right|\le{Kx\over\log^2x}
$$
As a result, using big O notation, the above thing becomes
$$
\begin{aligned}
\sum_{p\le x}\frac1p
&={1\over\log x}+\int_2^x{\mathrm dt\over t\log t}+\int_2^x\mathcal O\left(1\over t\log^2t\right)\mathrm dt+\mathcal O\left(1\over\log^2 x\right) \\
&=\log\log x+\mathcal O(1)
\end{aligned}
$$
That we can immediately conclude the error term is of $\mathcal O(1)$ is mainly because it is evident that the other terms would not exceed some constant.
P.S. Indeed a stronger result Mertens' theorem which states
$$
\sum_{p\le x}\frac1p=\log\log x+B_1+\mathcal O\left(1\over\log x\right)
$$
can be proven using elementary methods without prime number theorem, but it will require more technical details.
A: $$(1+O(\frac1{\log n}))n\log n = \log n! = \sum_{p^k \le n} \lfloor n/p^k \rfloor\log p $$ $$= (1+O(\frac1{\log n}))\sum_{p \le n} n \frac{\log p}{p}\tag{1}$$
Followed by a partial summation
$$\sum_{p\le n} \frac1p = \frac{\sum_{p\le n} \frac{\log p}p}{\log n}+\sum_{m \le n-1} (\sum_{p\le m} \frac{\log p}p) (\frac1{\log m}-\frac1{\log (m+1)})$$
$$ =  \frac{(1+O(\frac1{\log n}))\log n}{\log n}+\sum_{m\le n-1}(1+O(\frac1{\log m})) \log m  \frac{1+O(\frac1{m})}{m\log^2 m}$$ $$=\log \log n+C+O(\frac1{\log n})$$
