Proof of divergence of $1/2 + 1/3 + 1/5 + 1/7 + 1/11 +....$ What is an elementary proof to the fact that $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \dots$ diverges. ($p_i$ denotes the $i$th prime.)
 A: $$
\begin{align}
 \log \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \log \left( \prod_p \frac{1}{1-p^{-1}}\right)
  = \sum_p \log \left( \frac{1}{1-p^{-1}}\right) = \sum_p - \log(1-p^{-1}) \\
 & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
 & {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\
 & {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\
 & {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\
 & {} = \left( \sum_p \frac{1}{p} \right) + C
\end{align}$$
for a fixed constant $C<1.$ Since the harmonic series diverges, the sum of the reciprocals of the prime numbers diverges.
A: 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? ;)
