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 $K<1$?

You can in fact do better and show that $K < 0.43$.
$$
\sum_{p}\frac{1}{2p^2} +  \sum_{p}\frac{1}{3p
^3} +  \sum_{p}\frac{1}{4p^4} + \cdots = \sum_{n=1}^{\infty}\sum_{p}\frac{1}{np^n} 
$$
$$
<  \sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{nk^n} = \sum_{n=2}^{\infty}\frac{\zeta(n) - 1}{n} = 1 - \gamma \approx 0.422785
$$
where Gamma is the Euler-Mascheroni constant.
