I am trying to develop an argument that $\sum 1/p$ grows like $\log \log x$ for $p\leq x$ that is accessible to students without university training in mathematics.
I would like comment on its validity, especially the last step.
Step 1:
Let's start with the product formula.
$$\zeta(s)=\sum_{n}\frac{1}{n^{s}}=\prod_{n}\left(\frac{1}{1-p^{-s}}\right)$$
Turning products into sums by taking logarithms is a good habit.
$$\log\zeta(s)=-\sum_{p}\ln(1-p^{-s})$$
Step 2:
We can use the power series $\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\cdots$ for $|x|<1$ to expand the logarithm.
$$\begin{align}\ln\zeta(s)&=\sum_{p}(\frac{1}{p^{s}}+\frac{1}{2p^{2s}}+\frac{1}{3p^{3s}}+\frac{1}{4p^{4s}}\ldots) \\ &=\sum_{p}\frac{1}{p^{s}}+C\end{align}$$
The expansion gives us $\sum1/p^{s}$ which looks promising. The rest of the expansion converges to a constant $C$.
Step 3:
How do we know $C$ converges? We know $\sum\frac{1}{n^{x}}$, converges if $x>1$, and because $s>1$ this is also true for $\sum1/n^{ks}$ for $k\geq2$. Any series $\sum\frac{1}{p^{x}}$, summing only over primes p and not all integers $n$, also converges because it is a subseries of $\sum\frac{1}{n^{x}}$.
Step 4:
Let's see what happens if we set $s=1$. The rest of the series still converges to a constant, let's call it $D$, because the smallest $ks$ in $\sum1/p^{ks}$ is 2.
$$\ln\left(\sum\frac{1}{n}\right)=\sum_{p}\frac{1}{p}+D$$
Now, we know from integral comparison tests that $\sum1/n$ grows like $\ln(n)$, so $\ln\sum1/n$ must grow like $\ln\ln(n)$.
Since $D$ is a constant, this suggests $\sum1/p$ grows like $\ln\ln(n)$ for $p\leq n$.
Discussion:
The reason I am posting this question is that I think there is merit in step 4.
We know the LHS $\sum\frac{1}{n}$ diverges, but we also know that as $n$ grows, the sum grows like $\ln n$. We can then say that $\ln\left(\sum\frac{1}{n}\right)$ grows like $\ln \ln(n)$. Is this correct?
We know the RHS $\sum_{p}\frac{1}{p}+D$ diverges, but since it derived using the same mathematical operations as the LHS, it also grows in the same way as the LHS. Is this correct?
If the last statement is correct, because $D$ is known to converge, we can then conclude $\sum 1/p$ grows like $\ln \ln(n)$ for $p \leq n$.
There might be a question about the different ranges for the LHS and RHS. The LHS is up to $n$, and the RHS is up to $p$. I believe the argument can be made to equate them such that the common function domains is $p \leq n$. Is this correct?
I appreciate your patience with me on this, I hope you will value my persistence on this, and not giving up learning mathematics!
Note - I have asked a similar question before [link], but I feel this question is different and more specific.
Euler was wrong. Why do you write "grow like" instead of $\sum_{n\le x} \frac1n \sim \log x$? It implies that $\log(\sum_{n\le x} \frac1n) \sim \log\log x$ which doesn't tell anything about $\sum_{p\le x} \frac1p$.
There is only one elementary proof of the Mertens theorem $\sum_{p\le x} \frac1p \sim \log\log x$ whose main step is $$n\log n\sim \log n! =\sum_{p^k\le n} \lfloor n/p^k\rfloor\log p \sim n \sum_{p^k\le n}\frac{\log p}{p^k}$$ where the last step needs $\sum_{p^k\le n}\log p =O(n)$ which follows from $\sum_{p\in (n,2n]}\log p\le \log{2n\choose n}=O(n)$.
Followed by a partial summation to go from $\sum_{p\le x} \frac{\log p}{p}\sim\log x$ to $\sum_{p\le x} \frac1{p} \sim \log\log x$