Divergence of reciprocal of primes, Euler On Wikipedia at link currently is:
\begin{align}
 \ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
  = -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
 & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
 & {} = \sum_{p}\frac{1}{p}+  \sum_{p}\frac{1}{2p^2} +  \sum_{p}\frac{1}{3p
^3} +  \sum_{p}\frac{1}{4p^4} + \cdots  \\
 & {} = \left( \sum_p \frac{1}{p} \right) + K
\end{align}
And then Wikipedia says that $K<1$, without any explanation. How do we know that $$\sum_{p}\frac{1}{2p^2} +  \sum_{p}\frac{1}{3p
^3} +  \sum_{p}\frac{1}{4p^4} + \cdots$$ is equal to a constant $K<1$?

Hint : 
$$\sum_p \frac {1}{p^s} < \sum_{n=1}^{\infty} \frac 1{n^s} = \zeta(s) $$
$\zeta(s)$ is Riemann Zeta function, and $\zeta(s)$ converges for each $s \in \Bbb R, s>1$
