Proof of divergence of $1/2 + 1/3 + 1/5 + 1/7 + 1/11 +....$ [duplicate]

$$ \begin{align} \log \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \log \left( \prod_p \frac{1}{1-p^{-1}}\right) = \sum_p \log \left( \frac{1}{1-p^{-1}}\right) = \sum_p - \log(1-p^{-1}) \\ & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\ & {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\ & {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\ & {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\ & {} = \left( \sum_p \frac{1}{p} \right) + C \end{align}$$

for a fixed constant $C<1.$ Since the harmonic series diverges, the sum of the reciprocals of the prime numbers diverges.

This is an overkill, but since

$$\sum_{n=2}^\infty \frac{1}{n \ln(n)}$$

is divergent, it follows from the Prime Number Theorem that your series is divergent....

Now the question is: is the PNT considered an elemenatry tool? ;)